Polynomials

$\displaystyle \textbf{Question 1. }\text{Find the zeros of each of the following quadratic polynomials and }$
$\displaystyle \text{verify the relationship between the zeros and their coefficients.}$
$\displaystyle \textbf{(i) }f(x)=x^2-2x-8$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-2x-8$
$\displaystyle =x^2-4x+2x-8$
$\displaystyle =x(x-4)+2(x-4)$
$\displaystyle =(x-4)(x+2)$
$\displaystyle \text{The zeros of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (x-4)(x+2)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }x+2=0$
$\displaystyle \Rightarrow x=4\text{ or }x=-2$
$\displaystyle \text{Thus, }\alpha=4\text{ and }\beta=-2.$
$\displaystyle \alpha+\beta=4+(-2)=2,\quad \alpha\beta=4(-2)=-8\qquad ...(i)$
$\displaystyle \text{From }f(x)=x^2-2x-8,\text{ we find that}$
$\displaystyle \text{Coefficient of }x^2=1,\text{ Coefficient of }x=-2\text{ and Constant term}=-8$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{-2}{1}=2$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-8}{1}=-8\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(ii) }g(s)=4s^2-4s+1$
$\displaystyle \text{Answer:}$
$\displaystyle g(s)=4s^2-4s+1$
$\displaystyle =(2s-1)^2$
$\displaystyle \text{The zeros of }g(s)\text{ are given by }g(s)=0.$
$\displaystyle (2s-1)^2=0$
$\displaystyle \Rightarrow 2s-1=0$
$\displaystyle \Rightarrow s=\frac{1}{2}$
$\displaystyle \text{Thus, }\alpha=\frac{1}{2}\text{ and }\beta=\frac{1}{2}.$
$\displaystyle \alpha+\beta=\frac{1}{2}+\frac{1}{2}=1,\quad \alpha\beta=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\qquad ...(i)$
$\displaystyle \text{From }g(s)=4s^2-4s+1,\text{ we find that}$
$\displaystyle \text{Coefficient of }s^2=4,\text{ Coefficient of }s=-4\text{ and Constant term}=1$
$\displaystyle -\frac{\text{Coefficient of }s}{\text{Coefficient of }s^2}=-\frac{-4}{4}=1$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }s^2}=\frac{1}{4}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(iii) }h(t)=t^2-15\qquad [\mathrm{CBSE}\ 2024]$
$\displaystyle \text{Answer:}$
$\displaystyle h(t)=t^2-15$
$\displaystyle =t^2-(\sqrt{15})^2$
$\displaystyle =(t-\sqrt{15})(t+\sqrt{15})$
$\displaystyle \text{The zeros of }h(t)\text{ are given by }h(t)=0.$
$\displaystyle (t-\sqrt{15})(t+\sqrt{15})=0$
$\displaystyle \Rightarrow t-\sqrt{15}=0\text{ or }t+\sqrt{15}=0$
$\displaystyle \Rightarrow t=\sqrt{15}\text{ or }t=-\sqrt{15}$
$\displaystyle \text{Thus, }\alpha=\sqrt{15}\text{ and }\beta=-\sqrt{15}.$
$\displaystyle \alpha+\beta=\sqrt{15}+(-\sqrt{15})=0,\quad \alpha\beta=\sqrt{15}(-\sqrt{15})=-15\qquad ...(i)$
$\displaystyle \text{From }h(t)=t^2+0t-15,\text{ we find that}$
$\displaystyle \text{Coefficient of }t^2=1,\text{ Coefficient of }t=0\text{ and Constant term}=-15$
$\displaystyle -\frac{\text{Coefficient of }t}{\text{Coefficient of }t^2}=-\frac{0}{1}=0$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }t^2}=\frac{-15}{1}=-15\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(iv) }f(x)=6x^2-3-7x$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=6x^2-3-7x=6x^2-7x-3$
$\displaystyle =6x^2-9x+2x-3$
$\displaystyle =3x(2x-3)+1(2x-3)$
$\displaystyle =(2x-3)(3x+1)$
$\displaystyle \text{The zeros of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (2x-3)(3x+1)=0$
$\displaystyle \Rightarrow 2x-3=0\text{ or }3x+1=0$
$\displaystyle \Rightarrow x=\frac{3}{2}\text{ or }x=-\frac{1}{3}$
$\displaystyle \text{Thus, }\alpha=\frac{3}{2}\text{ and }\beta=-\frac{1}{3}.$
$\displaystyle \alpha+\beta=\frac{3}{2}+\left(-\frac{1}{3}\right)=\frac{9-2}{6}=\frac{7}{6}$
$\displaystyle \alpha\beta=\frac{3}{2}\times\left(-\frac{1}{3}\right)=-\frac{1}{2}\qquad ...(i)$
$\displaystyle \text{From }f(x)=6x^2-7x-3,\text{ we find that}$
$\displaystyle \text{Coefficient of }x^2=6,\text{ Coefficient of }x=-7\text{ and Constant term}=-3$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{-7}{6}=\frac{7}{6}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-3}{6}=-\frac{1}{2}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(v) }q(y)=7y^2-\frac{11}{3}y-\frac{2}{3}\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle q(y)=7y^2-\frac{11}{3}y-\frac{2}{3}$
$\displaystyle =\frac{1}{3}(21y^2-11y-2)$
$\displaystyle =\frac{1}{3}(21y^2-14y+3y-2)$
$\displaystyle =\frac{1}{3}\{7y(3y-2)+1(3y-2)\}$
$\displaystyle =\frac{1}{3}(3y-2)(7y+1)$
$\displaystyle \text{The zeros of }q(y)\text{ are given by }q(y)=0.$
$\displaystyle \frac{1}{3}(3y-2)(7y+1)=0$
$\displaystyle \Rightarrow 3y-2=0\text{ or }7y+1=0$
$\displaystyle \Rightarrow y=\frac{2}{3}\text{ or }y=-\frac{1}{7}$
$\displaystyle \text{Thus, }\alpha=\frac{2}{3}\text{ and }\beta=-\frac{1}{7}.$
$\displaystyle \alpha+\beta=\frac{2}{3}+\left(-\frac{1}{7}\right)=\frac{14-3}{21}=\frac{11}{21}$
$\displaystyle \alpha\beta=\frac{2}{3}\times\left(-\frac{1}{7}\right)=-\frac{2}{21}\qquad ...(i)$
$\displaystyle \text{From }q(y)=7y^2-\frac{11}{3}y-\frac{2}{3},\text{ we find that}$
$\displaystyle \text{Coefficient of }y^2=7,\text{ Coefficient of }y=-\frac{11}{3}\text{ and Constant term}=-\frac{2}{3}$
$\displaystyle -\frac{\text{Coefficient of }y}{\text{Coefficient of }y^2}=-\frac{-\frac{11}{3}}{7}=\frac{11}{21}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }y^2}=\frac{-\frac{2}{3}}{7}=-\frac{2}{21}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(vi) }\phi(x)=2x^2+\frac{7}{2}x+\frac{3}{4}$
$\displaystyle \text{Answer:}$
$\displaystyle \phi(x)=2x^2+\frac{7}{2}x+\frac{3}{4}$
$\displaystyle =\frac{1}{4}(8x^2+14x+3)$
$\displaystyle =\frac{1}{4}(8x^2+12x+2x+3)$
$\displaystyle =\frac{1}{4}\{4x(2x+3)+1(2x+3)\}$
$\displaystyle =\frac{1}{4}(2x+3)(4x+1)$
$\displaystyle \text{The zeros of }\phi(x)\text{ are given by }\phi(x)=0.$
$\displaystyle \frac{1}{4}(2x+3)(4x+1)=0$
$\displaystyle \Rightarrow 2x+3=0\text{ or }4x+1=0$
$\displaystyle \Rightarrow x=-\frac{3}{2}\text{ or }x=-\frac{1}{4}$
$\displaystyle \text{Thus, }\alpha=-\frac{3}{2}\text{ and }\beta=-\frac{1}{4}.$
$\displaystyle \alpha+\beta=-\frac{3}{2}+\left(-\frac{1}{4}\right)=\frac{-6-1}{4}=-\frac{7}{4}$
$\displaystyle \alpha\beta=\left(-\frac{3}{2}\right)\left(-\frac{1}{4}\right)=\frac{3}{8}\qquad ...(i)$
$\displaystyle \text{From }\phi(x)=2x^2+\frac{7}{2}x+\frac{3}{4},\text{ we find that}$
$\displaystyle \text{Coefficient of }x^2=2,\text{ Coefficient of }x=\frac{7}{2}\text{ and Constant term}=\frac{3}{4}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{\frac{7}{2}}{2}=-\frac{7}{4}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{\frac{3}{4}}{2}=\frac{3}{8}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(vii) }p(x)=x^2+\frac{4}{3}x-\frac{4}{3}\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle p(x)=x^2+\frac{4}{3}x-\frac{4}{3}$
$\displaystyle =\frac{1}{3}(3x^2+4x-4)$
$\displaystyle =\frac{1}{3}(3x^2+6x-2x-4)$
$\displaystyle =\frac{1}{3}\{3x(x+2)-2(x+2)\}$
$\displaystyle =\frac{1}{3}(x+2)(3x-2)$
$\displaystyle \text{The zeros of }p(x)\text{ are given by }p(x)=0.$
$\displaystyle \frac{1}{3}(x+2)(3x-2)=0$
$\displaystyle \Rightarrow x+2=0\text{ or }3x-2=0$
$\displaystyle \Rightarrow x=-2\text{ or }x=\frac{2}{3}$
$\displaystyle \text{Thus, }\alpha=-2\text{ and }\beta=\frac{2}{3}.$
$\displaystyle \alpha+\beta=-2+\frac{2}{3}=\frac{-6+2}{3}=-\frac{4}{3}$
$\displaystyle \alpha\beta=-2\times\frac{2}{3}=-\frac{4}{3}\qquad ...(i)$
$\displaystyle \text{From }p(x)=x^2+\frac{4}{3}x-\frac{4}{3},\text{ we find that}$
$\displaystyle \text{Coefficient of }x^2=1,\text{ Coefficient of }x=\frac{4}{3}\text{ and Constant term}=-\frac{4}{3}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{\frac{4}{3}}{1}=-\frac{4}{3}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-\frac{4}{3}}{1}=-\frac{4}{3}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{Question 2. }\text{For each of the following, find a quadratic polynomial whose sum}$
$\displaystyle \text{and product respectively of the zeroes are as given. Also, find the zeroes of these} \\ \text{polynomials by factorization.}$

$\displaystyle \textbf{(i) }\text{Sum}=-\frac{8}{3},\text{ Product}=\frac{4}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that a quadratic polynomial when the sum and product of its zeroes are given is}$
$\displaystyle f(x)=k\{x^2-(\text{Sum of the zeroes})x+\text{Product of the zeroes}\}$
$\displaystyle f(x)=k\left\{x^2-\left(-\frac{8}{3}\right)x+\frac{4}{3}\right\}$
$\displaystyle =k\left(x^2+\frac{8}{3}x+\frac{4}{3}\right)$
$\displaystyle =\frac{k}{3}(3x^2+8x+4)$
$\displaystyle \text{Taking }\frac{k}{3}=1,\text{ we get}$
$\displaystyle f(x)=3x^2+8x+4$
$\displaystyle =3x^2+6x+2x+4$
$\displaystyle =3x(x+2)+2(x+2)$
$\displaystyle =(x+2)(3x+2)$
$\displaystyle \text{The zeroes of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (x+2)(3x+2)=0$
$\displaystyle \Rightarrow x+2=0\text{ or }3x+2=0$
$\displaystyle \Rightarrow x=-2\text{ or }x=-\frac{2}{3}$
$\displaystyle \text{Hence, the required polynomial is }3x^2+8x+4\text{ and its zeroes are }-2\text{ and }-\frac{2}{3}.$
$\\$

$\displaystyle \textbf{(ii) }\text{Sum}=\frac{21}{8},\text{ Product}=\frac{5}{16}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=k\{x^2-(\text{Sum of the zeroes})x+\text{Product of the zeroes}\}$
$\displaystyle f(x)=k\left(x^2-\frac{21}{8}x+\frac{5}{16}\right)$
$\displaystyle =\frac{k}{16}(16x^2-42x+5)$
$\displaystyle \text{Taking }\frac{k}{16}=1,\text{ we get}$
$\displaystyle f(x)=16x^2-42x+5$
$\displaystyle =16x^2-40x-2x+5$
$\displaystyle =8x(2x-5)-1(2x-5)$
$\displaystyle =(2x-5)(8x-1)$
$\displaystyle \text{The zeroes of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (2x-5)(8x-1)=0$
$\displaystyle \Rightarrow 2x-5=0\text{ or }8x-1=0$
$\displaystyle \Rightarrow x=\frac{5}{2}\text{ or }x=\frac{1}{8}$
$\displaystyle \text{Hence, the required polynomial is }16x^2-42x+5\text{ and its zeroes are }\frac{5}{2}\text{ and }\frac{1}{8}.$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial }$
$\displaystyle f(x)=x^2-5x+4, \ \text{find the value of }\frac{1}{\alpha}+\frac{1}{\beta}-2\alpha\beta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }f(x)=x^2-5x+4.$
$\displaystyle \therefore \alpha+\beta=-\frac{-5}{1}=5,\quad \alpha\beta=\frac{4}{1}=4$
$\displaystyle \text{Now,}$
$\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}-2\alpha\beta=\frac{\alpha+\beta}{\alpha\beta}-2\alpha\beta$
$\displaystyle =\frac{5}{4}-2(4)$
$\displaystyle =\frac{5}{4}-8$
$\displaystyle =\frac{5-32}{4}$
$\displaystyle =-\frac{27}{4}$
$\displaystyle \text{Hence, }\frac{1}{\alpha}+\frac{1}{\beta}-2\alpha\beta=-\frac{27}{4}.$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial }$
$\displaystyle p(y)=5y^2-7y+1, \ \text{find the value of }\frac{1}{\alpha}+\frac{1}{\beta}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }p(y)=5y^2-7y+1.$
$\displaystyle \therefore \alpha+\beta=-\frac{-7}{5}=\frac{7}{5},\quad \alpha\beta=\frac{1}{5}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}$
$\displaystyle =\frac{\frac{7}{5}}{\frac{1}{5}}$
$\displaystyle =7$
$\displaystyle \text{Hence, }\frac{1}{\alpha}+\frac{1}{\beta}=7.$
$\\$

$\displaystyle \textbf{Question 5. }\text{If one zero of the quadratic polynomial }$
$\displaystyle f(x)=4x^2-8kx-9\text{ is negative of the other,} \ \text{find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha\text{ and }\beta\text{ be the zeroes of }f(x)=4x^2-8kx-9.$
$\displaystyle \therefore \alpha+\beta=-\frac{-8k}{4}=2k$
$\displaystyle \text{It is given that one zero is negative of the other.}$
$\displaystyle \therefore \alpha+\beta=0$
$\displaystyle \Rightarrow 2k=0$
$\displaystyle \Rightarrow k=0$
$\displaystyle \text{Hence, }k=0.$
$\\$

$\displaystyle \textbf{Question 6. }\text{If the sum of the zeroes of the quadratic polynomial }$
$\displaystyle f(t)=kt^2+2t+3k\text{ is equal to their} \ \text{product, find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha\text{ and }\beta\text{ be the zeroes of }f(t)=kt^2+2t+3k.$
$\displaystyle \therefore \alpha+\beta=-\frac{2}{k},\quad \alpha\beta=\frac{3k}{k}=3$
$\displaystyle \text{It is given that the sum of the zeroes is equal to their product.}$
$\displaystyle \therefore \alpha+\beta=\alpha\beta$
$\displaystyle \Rightarrow -\frac{2}{k}=3$
$\displaystyle \Rightarrow 3k=-2$
$\displaystyle \Rightarrow k=-\frac{2}{3}$
$\displaystyle \text{Hence, }k=-\frac{2}{3}.$
$\\$

$\displaystyle \textbf{Question 7. }\text{Find the zeros of each of the following quadratic polynomials and }$
$\displaystyle \text{verify the relationship between the zeros and their coefficients.}$

$\displaystyle \textbf{(i) }p(x)=x^2+2\sqrt{2}x-6$
$\displaystyle \text{Answer:}$
$\displaystyle p(x)=x^2+2\sqrt{2}x-6$
$\displaystyle =x^2+3\sqrt{2}x-\sqrt{2}x-6$
$\displaystyle =x(x+3\sqrt{2})-\sqrt{2}(x+3\sqrt{2})$
$\displaystyle =(x+3\sqrt{2})(x-\sqrt{2})$
$\displaystyle \text{The zeros of }p(x)\text{ are given by }p(x)=0.$
$\displaystyle (x+3\sqrt{2})(x-\sqrt{2})=0$
$\displaystyle \Rightarrow x=-3\sqrt{2}\text{ or }x=\sqrt{2}$
$\displaystyle \text{Thus, }\alpha=-3\sqrt{2}\text{ and }\beta=\sqrt{2}.$
$\displaystyle \alpha+\beta=-3\sqrt{2}+\sqrt{2}=-2\sqrt{2},\quad \alpha\beta=(-3\sqrt{2})(\sqrt{2})=-6\qquad ...(i)$
$\displaystyle \text{From }p(x)=x^2+2\sqrt{2}x-6,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{2\sqrt{2}}{1}=-2\sqrt{2}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-6}{1}=-6\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(ii) }q(x)=\sqrt{3}x^2+10x+7\sqrt{3}$
$\displaystyle \text{Answer:}$
$\displaystyle q(x)=\sqrt{3}x^2+10x+7\sqrt{3}$
$\displaystyle =\sqrt{3}x^2+3x+7x+7\sqrt{3}$
$\displaystyle =x(\sqrt{3}x+3)+7(x+\sqrt{3})$
$\displaystyle =(x+\sqrt{3})(\sqrt{3}x+7)$
$\displaystyle q(x)=0\Rightarrow (x+\sqrt{3})(\sqrt{3}x+7)=0$
$\displaystyle \Rightarrow x=-\sqrt{3}\text{ or }x=-\frac{7}{\sqrt{3}}$
$\displaystyle \text{Thus, }\alpha=-\sqrt{3}\text{ and }\beta=-\frac{7}{\sqrt{3}}.$
$\displaystyle \alpha+\beta=-\sqrt{3}-\frac{7}{\sqrt{3}}=-\frac{10}{\sqrt{3}},\quad \alpha\beta=(-\sqrt{3})\left(-\frac{7}{\sqrt{3}}\right)=7\qquad ...(i)$
$\displaystyle \text{From }q(x)=\sqrt{3}x^2+10x+7\sqrt{3},\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{10}{\sqrt{3}},\quad \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{7\sqrt{3}}{\sqrt{3}}=7\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(iii) }f(x)=x^2-(\sqrt{3}+1)x+\sqrt{3}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-(\sqrt{3}+1)x+\sqrt{3}$
$\displaystyle =x^2-\sqrt{3}x-x+\sqrt{3}$
$\displaystyle =x(x-\sqrt{3})-1(x-\sqrt{3})$
$\displaystyle =(x-\sqrt{3})(x-1)$
$\displaystyle f(x)=0\Rightarrow (x-\sqrt{3})(x-1)=0$
$\displaystyle \Rightarrow x=\sqrt{3}\text{ or }x=1$
$\displaystyle \text{Thus, }\alpha=\sqrt{3}\text{ and }\beta=1.$
$\displaystyle \alpha+\beta=\sqrt{3}+1,\quad \alpha\beta=\sqrt{3}\qquad ...(i)$
$\displaystyle \text{From }f(x)=x^2-(\sqrt{3}+1)x+\sqrt{3},\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{-(\sqrt{3}+1)}{1}=\sqrt{3}+1$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{\sqrt{3}}{1}=\sqrt{3}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(iv) }g(x)=a(x^2+1)-x(a^2+1)$
$\displaystyle \text{Answer:}$
$\displaystyle g(x)=a(x^2+1)-x(a^2+1)$
$\displaystyle =ax^2+a-a^2x-x$
$\displaystyle =ax^2-a^2x-x+a$
$\displaystyle =ax(x-a)-1(x-a)$
$\displaystyle =(x-a)(ax-1)$
$\displaystyle g(x)=0\Rightarrow (x-a)(ax-1)=0$
$\displaystyle \Rightarrow x=a\text{ or }x=\frac{1}{a}$
$\displaystyle \text{Thus, }\alpha=a\text{ and }\beta=\frac{1}{a}.$
$\displaystyle \alpha+\beta=a+\frac{1}{a}=\frac{a^2+1}{a},\quad \alpha\beta=a\cdot\frac{1}{a}=1\qquad ...(i)$
$\displaystyle \text{From }g(x)=ax^2-(a^2+1)x+a,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{-(a^2+1)}{a}=\frac{a^2+1}{a}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{a}{a}=1\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(v) }h(s)=2s^2-(1+2\sqrt{2})s+\sqrt{2}$
$\displaystyle \text{Answer:}$
$\displaystyle h(s)=2s^2-(1+2\sqrt{2})s+\sqrt{2}$
$\displaystyle =2s^2-s-2\sqrt{2}s+\sqrt{2}$
$\displaystyle =s(2s-1)-\sqrt{2}(2s-1)$
$\displaystyle =(2s-1)(s-\sqrt{2})$
$\displaystyle h(s)=0\Rightarrow (2s-1)(s-\sqrt{2})=0$
$\displaystyle \Rightarrow s=\frac{1}{2}\text{ or }s=\sqrt{2}$
$\displaystyle \text{Thus, }\alpha=\frac{1}{2}\text{ and }\beta=\sqrt{2}.$
$\displaystyle \alpha+\beta=\frac{1}{2}+\sqrt{2}=\frac{1+2\sqrt{2}}{2},\quad \alpha\beta=\frac{\sqrt{2}}{2}\qquad ...(i)$
$\displaystyle \text{From }h(s)=2s^2-(1+2\sqrt{2})s+\sqrt{2},\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }s}{\text{Coefficient of }s^2}=-\frac{-(1+2\sqrt{2})}{2}=\frac{1+2\sqrt{2}}{2}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }s^2}=\frac{\sqrt{2}}{2}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(vi) }f(v)=v^2+4\sqrt{3}v-15$
$\displaystyle \text{Answer:}$
$\displaystyle f(v)=v^2+4\sqrt{3}v-15$
$\displaystyle =v^2+5\sqrt{3}v-\sqrt{3}v-15$
$\displaystyle =v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})$
$\displaystyle =(v+5\sqrt{3})(v-\sqrt{3})$
$\displaystyle f(v)=0\Rightarrow (v+5\sqrt{3})(v-\sqrt{3})=0$
$\displaystyle \Rightarrow v=-5\sqrt{3}\text{ or }v=\sqrt{3}$
$\displaystyle \text{Thus, }\alpha=-5\sqrt{3}\text{ and }\beta=\sqrt{3}.$
$\displaystyle \alpha+\beta=-5\sqrt{3}+\sqrt{3}=-4\sqrt{3},\quad \alpha\beta=(-5\sqrt{3})(\sqrt{3})=-15\qquad ...(i)$
$\displaystyle \text{From }f(v)=v^2+4\sqrt{3}v-15,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }v}{\text{Coefficient of }v^2}=-\frac{4\sqrt{3}}{1}=-4\sqrt{3}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }v^2}=\frac{-15}{1}=-15\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(vii) }p(y)=y^2+\frac{3\sqrt{5}}{2}y-5$
$\displaystyle \text{Answer:}$
$\displaystyle p(y)=y^2+\frac{3\sqrt{5}}{2}y-5$
$\displaystyle =y^2+2\sqrt{5}y-\frac{\sqrt{5}}{2}y-5$
$\displaystyle =y(y+2\sqrt{5})-\frac{\sqrt{5}}{2}(y+2\sqrt{5})$
$\displaystyle =(y+2\sqrt{5})\left(y-\frac{\sqrt{5}}{2}\right)$
$\displaystyle p(y)=0\Rightarrow (y+2\sqrt{5})\left(y-\frac{\sqrt{5}}{2}\right)=0$
$\displaystyle \Rightarrow y=-2\sqrt{5}\text{ or }y=\frac{\sqrt{5}}{2}$
$\displaystyle \text{Thus, }\alpha=-2\sqrt{5}\text{ and }\beta=\frac{\sqrt{5}}{2}.$
$\displaystyle \alpha+\beta=-2\sqrt{5}+\frac{\sqrt{5}}{2}=-\frac{3\sqrt{5}}{2},\quad \alpha\beta=(-2\sqrt{5})\left(\frac{\sqrt{5}}{2}\right)=-5\qquad ...(i)$
$\displaystyle \text{From }p(y)=y^2+\frac{3\sqrt{5}}{2}y-5,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }y}{\text{Coefficient of }y^2}=-\frac{\frac{3\sqrt{5}}{2}}{1}=-\frac{3\sqrt{5}}{2}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }y^2}=\frac{-5}{1}=-5\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(viii) }\phi(x)=4x^2+5\sqrt{2}x-3$
$\displaystyle \text{Answer:}$
$\displaystyle \phi(x)=4x^2+5\sqrt{2}x-3$
$\displaystyle =4x^2+6\sqrt{2}x-\sqrt{2}x-3$
$\displaystyle =2x(2x+3\sqrt{2})-\frac{\sqrt{2}}{2}(2x+3\sqrt{2})$
$\displaystyle =\left(2x+3\sqrt{2}\right)\left(2x-\frac{\sqrt{2}}{2}\right)$
$\displaystyle \phi(x)=0\Rightarrow \left(2x+3\sqrt{2}\right)\left(2x-\frac{\sqrt{2}}{2}\right)=0$
$\displaystyle \Rightarrow x=-\frac{3\sqrt{2}}{2}\text{ or }x=\frac{\sqrt{2}}{4}$
$\displaystyle \text{Thus, }\alpha=-\frac{3\sqrt{2}}{2}\text{ and }\beta=\frac{\sqrt{2}}{4}.$
$\displaystyle \alpha+\beta=-\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{4}=-\frac{5\sqrt{2}}{4},\quad \alpha\beta=-\frac{3}{4}\qquad ...(i)$
$\displaystyle \text{From }\phi(x)=4x^2+5\sqrt{2}x-3,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{5\sqrt{2}}{4}$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-3}{4}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(ix) }f(x)=4x^2+4x-3\qquad [\mathrm{CBSE}\ 2024]$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=4x^2+4x-3$
$\displaystyle =4x^2+6x-2x-3$
$\displaystyle =2x(2x+3)-1(2x+3)$
$\displaystyle =(2x+3)(2x-1)$
$\displaystyle f(x)=0\Rightarrow (2x+3)(2x-1)=0$
$\displaystyle \Rightarrow x=-\frac{3}{2}\text{ or }x=\frac{1}{2}$
$\displaystyle \text{Thus, }\alpha=-\frac{3}{2}\text{ and }\beta=\frac{1}{2}.$
$\displaystyle \alpha+\beta=-\frac{3}{2}+\frac{1}{2}=-1,\quad \alpha\beta=-\frac{3}{4}\qquad ...(i)$
$\displaystyle \text{From }f(x)=4x^2+4x-3,\text{ we find that}$
$\displaystyle -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}=-\frac{4}{4}=-1$
$\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }x^2}=\frac{-3}{4}\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{Question 8. }\text{Find a quadratic polynomial whose sum and product respectively}$
$\displaystyle \text{of the zeroes are given. Also, find the zeroes of these polynomials.}\qquad [\mathrm{CBSE}\ 2024]$

$\displaystyle \textbf{(i) }\text{Sum}=-2\sqrt{3},\text{ Product}=-9$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that a quadratic polynomial when the sum and product of its zeroes are given is}$
$\displaystyle f(x)=k\left\{x^2-(\text{Sum of the zeroes})x+\text{Product of the zeroes}\right\}$
$\displaystyle f(x)=k\left(x^2+2\sqrt{3}x-9\right)$
$\displaystyle =k\left(x^2+3\sqrt{3}x-\sqrt{3}x-9\right)$
$\displaystyle =k\left\{x(x+3\sqrt{3})-\sqrt{3}(x+3\sqrt{3})\right\}$
$\displaystyle =k(x+3\sqrt{3})(x-\sqrt{3})$
$\displaystyle \text{Taking }k=1,\text{ the required polynomial is}$
$\displaystyle f(x)=x^2+2\sqrt{3}x-9$
$\displaystyle \text{The zeroes of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (x+3\sqrt{3})(x-\sqrt{3})=0$
$\displaystyle \Rightarrow x=-3\sqrt{3}\text{ or }x=\sqrt{3}$
$\displaystyle \text{Hence, the required polynomial is }x^2+2\sqrt{3}x-9\text{ and its zeroes are }-3\sqrt{3}\text{ and }\sqrt{3}.$
$\\$

$\displaystyle \textbf{(ii) }\text{Sum}=-\frac{3}{2\sqrt{5}},\text{ Product}=-\frac{1}{2}\qquad [\mathrm{CBSE}\ 2024]$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=k\left\{x^2-(\text{Sum of the zeroes})x+\text{Product of the zeroes}\right\}$
$\displaystyle f(x)=k\left(x^2+\frac{3}{2\sqrt{5}}x-\frac{1}{2}\right)$
$\displaystyle =\frac{k}{2}\left(2x^2+\frac{3}{\sqrt{5}}x-1\right)$
$\displaystyle =\frac{k}{2\sqrt{5}}\left(2\sqrt{5}x^2+3x-\sqrt{5}\right)$
$\displaystyle =\frac{k}{2\sqrt{5}}\left(2\sqrt{5}x^2+5x-2x-\sqrt{5}\right)$
$\displaystyle =\frac{k}{2\sqrt{5}}\left\{\sqrt{5}x(2x+\sqrt{5})-1(2x+\sqrt{5})\right\}$
$\displaystyle =\frac{k}{2\sqrt{5}}(2x+\sqrt{5})(\sqrt{5}x-1)$
$\displaystyle \text{Taking }\frac{k}{2\sqrt{5}}=1,\text{ the required polynomial is}$
$\displaystyle f(x)=2\sqrt{5}x^2+3x-\sqrt{5}$
$\displaystyle \text{The zeroes of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (2x+\sqrt{5})(\sqrt{5}x-1)=0$
$\displaystyle \Rightarrow 2x+\sqrt{5}=0\text{ or }\sqrt{5}x-1=0$
$\displaystyle \Rightarrow x=-\frac{\sqrt{5}}{2}\text{ or }x=\frac{1}{\sqrt{5}}$
$\displaystyle \text{Hence, the required polynomial is }2\sqrt{5}x^2+3x-\sqrt{5}\text{ and its zeroes are }-\frac{\sqrt{5}}{2}\text{ and }\frac{1}{\sqrt{5}}.$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial }$
$\displaystyle p(x)=4x^2-5x-1,\ \text{find the value of }\alpha^2\beta+\alpha\beta^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }p(x)=4x^2-5x-1.$
$\displaystyle \therefore \alpha+\beta=-\frac{-5}{4}=\frac{5}{4},\quad \alpha\beta=\frac{-1}{4}$
$\displaystyle \text{Now,}$
$\displaystyle \alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)$
$\displaystyle =-\frac{1}{4}\times\frac{5}{4}$
$\displaystyle =-\frac{5}{16}$
$\displaystyle \text{Hence, }\alpha^2\beta+\alpha\beta^2=-\frac{5}{16}.$
$\\$

$\displaystyle \textbf{Question 10. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial }$
$\displaystyle f(x)=6x^2+x-2, \ \text{find the value of }\frac{\alpha}{\beta}+\frac{\beta}{\alpha}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }f(x)=6x^2+x-2.$
$\displaystyle \therefore \alpha+\beta=-\frac{1}{6},\quad \alpha\beta=\frac{-2}{6}=-\frac{1}{3}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}$
$\displaystyle =\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
$\displaystyle =\frac{\left(-\frac{1}{6}\right)^2-2\left(-\frac{1}{3}\right)}{-\frac{1}{3}}$
$\displaystyle =\frac{\frac{1}{36}+\frac{2}{3}}{-\frac{1}{3}}$
$\displaystyle =\frac{\frac{1+24}{36}}{-\frac{1}{3}}$
$\displaystyle =\frac{\frac{25}{36}}{-\frac{1}{3}}$
$\displaystyle =-\frac{25}{12}$
$\displaystyle \text{Hence, }\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=-\frac{25}{12}.$
$\\$

$\displaystyle \textbf{Question 11. }\text{If }\alpha\text{ and }\beta\text{ are the zeros of the quadratic polynomial }$
$\displaystyle f(x)=x^2-p(x+1)-c, \ \text{show that }(\alpha+1)(\beta+1)=1-c.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-p(x+1)-c$
$\displaystyle =x^2-px-p-c$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeros of }f(x)=x^2-px-(p+c).$
$\displaystyle \therefore \alpha+\beta=-\frac{-p}{1}=p,\quad \alpha\beta=\frac{-(p+c)}{1}=-(p+c)$
$\displaystyle \text{Now,}$
$\displaystyle (\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1$
$\displaystyle =-(p+c)+p+1$
$\displaystyle =1-c$
$\displaystyle \text{Hence, }(\alpha+1)(\beta+1)=1-c.$
$\\$

$\displaystyle \textbf{Question 12. }\text{If }\alpha\text{ and }\beta\text{ are the zeros of a quadratic polynomial such that }$
$\displaystyle \alpha+\beta=24 \ \text{and }\alpha-\beta=8,\text{ find a quadratic polynomial having }\alpha\text{ and }\beta\text{ as its zeros.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that}$
$\displaystyle \alpha+\beta=24\qquad ...(i)$
$\displaystyle \alpha-\beta=8\qquad ...(ii)$
$\displaystyle \text{Adding (i) and (ii), we get}$
$\displaystyle 2\alpha=32$
$\displaystyle \Rightarrow \alpha=16$
$\displaystyle \text{Substituting }\alpha=16\text{ in (i), we get}$
$\displaystyle 16+\beta=24$
$\displaystyle \Rightarrow \beta=8$
$\displaystyle \therefore \alpha+\beta=24,\quad \alpha\beta=16\times 8=128$
$\displaystyle \text{Required quadratic polynomial is}$
$\displaystyle f(x)=k\{x^2-(\alpha+\beta)x+\alpha\beta\}$
$\displaystyle =k(x^2-24x+128)$
$\displaystyle \text{Taking }k=1,\text{ the required quadratic polynomial is }x^2-24x+128.$
$\\$

$\displaystyle \textbf{Question 13. }\text{Find a quadratic polynomial whose zeros are negative of the zeros of} \\ \text{the polynomial }px^2+qx+r.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha\text{ and }\beta\text{ be the zeros of the polynomial }px^2+qx+r.$
$\displaystyle \therefore \alpha+\beta=-\frac{q}{p},\quad \alpha\beta=\frac{r}{p}$
$\displaystyle \text{Zeros of the required polynomial are }-\alpha\text{ and }-\beta.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeros of the required polynomial.}$
$\displaystyle S=(-\alpha)+(-\beta)=-(\alpha+\beta)=-\left(-\frac{q}{p}\right)=\frac{q}{p}$
$\displaystyle P=(-\alpha)(-\beta)=\alpha\beta=\frac{r}{p}$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left(x^2-\frac{q}{p}x+\frac{r}{p}\right)$
$\displaystyle =\frac{k}{p}(px^2-qx+r)$
$\displaystyle \text{Taking }\frac{k}{p}=1,\text{ the required polynomial is }px^2-qx+r.$
$\\$

$\displaystyle \textbf{Question 14. }\text{If }\alpha\text{ and }\beta\text{ are zeroes of a quadratic polynomial }px^2+qx+1.$
$\displaystyle \text{Form a quadratic polynomial whose zeroes are }\frac{2}{\alpha}\text{ and }\frac{2}{\beta}.\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are zeroes of }px^2+qx+1.$
$\displaystyle \therefore \alpha+\beta=-\frac{q}{p},\quad \alpha\beta=\frac{1}{p}$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle \text{Then,}$
$\displaystyle S=\frac{2}{\alpha}+\frac{2}{\beta}=\frac{2(\alpha+\beta)}{\alpha\beta}$
$\displaystyle =\frac{2\left(-\frac{q}{p}\right)}{\frac{1}{p}}=-2q$
$\displaystyle \text{and,}$
$\displaystyle P=\frac{2}{\alpha}\times\frac{2}{\beta}=\frac{4}{\alpha\beta}$
$\displaystyle =\frac{4}{\frac{1}{p}}=4p$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\{x^2-(-2q)x+4p\}$
$\displaystyle =k(x^2+2qx+4p),\text{ where }k\text{ is any non-zero constant.}$
$\\$

$\displaystyle \textbf{Question 15. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the polynomial }ax^2-x+c.$
$\displaystyle \text{Obtain a polynomial whose zeroes are }\alpha-3\text{ and }\beta-3.\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }ax^2-x+c.$
$\displaystyle \therefore \alpha+\beta=-\frac{-1}{a}=\frac{1}{a},\quad \alpha\beta=\frac{c}{a}$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle \text{Then,}$
$\displaystyle S=(\alpha-3)+(\beta-3)=\alpha+\beta-6=\frac{1}{a}-6=\frac{1-6a}{a}$
$\displaystyle \text{and,}$
$\displaystyle P=(\alpha-3)(\beta-3)$
$\displaystyle =\alpha\beta-3\alpha-3\beta+9$
$\displaystyle =\alpha\beta-3(\alpha+\beta)+9$
$\displaystyle =\frac{c}{a}-3\left(\frac{1}{a}\right)+9$
$\displaystyle =\frac{c-3+9a}{a}$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left\{x^2-\frac{1-6a}{a}x+\frac{c-3+9a}{a}\right\}$
$\displaystyle =\frac{k}{a}\left\{ax^2-(1-6a)x+c-3+9a\right\}$
$\displaystyle =\frac{k}{a}\left\{ax^2+(6a-1)x+c+9a-3\right\}$
$\displaystyle \text{Taking }\frac{k}{a}=1,\text{ the required polynomial is}$
$\displaystyle ax^2+(6a-1)x+c+9a-3.$
$\\$

$\displaystyle \textbf{Question 16. }\text{Obtain the zeroes of the polynomial }7x^2+18x-9.\text{ Hence, write a}$
$\displaystyle \text{polynomial each of whose zeroes is twice the zeroes of the given polynomial.}\ [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=7x^2+18x-9$
$\displaystyle =7x^2+21x-3x-9$
$\displaystyle =7x(x+3)-3(x+3)$
$\displaystyle =(x+3)(7x-3)$
$\displaystyle \text{The zeroes of }f(x)\text{ are given by }f(x)=0.$
$\displaystyle (x+3)(7x-3)=0$
$\displaystyle \Rightarrow x=-3\text{ or }x=\frac{3}{7}$
$\displaystyle \text{Thus, }\alpha=-3\text{ and }\beta=\frac{3}{7}.$
$\displaystyle \text{The zeroes of the required polynomial are }2\alpha=-6\text{ and }2\beta=\frac{6}{7}.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle S=-6+\frac{6}{7}=-\frac{36}{7}$
$\displaystyle P=(-6)\times\frac{6}{7}=-\frac{36}{7}$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left(x^2+\frac{36}{7}x-\frac{36}{7}\right)$
$\displaystyle =\frac{k}{7}(7x^2+36x-36)$
$\displaystyle \text{Taking }\frac{k}{7}=1,\text{ the required polynomial is}$
$\displaystyle 7x^2+36x-36.$
$\\$

$\displaystyle \textbf{Question 17. }\text{Obtain the zeroes of the polynomial }p(x)=2x^2-5x-3.\text{ Hence, obtain a}$
$\displaystyle \text{polynomial each of whose zeroes is one less than each of the zeroes of }p(x).\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle p(x)=2x^2-5x-3$
$\displaystyle =2x^2-6x+x-3$
$\displaystyle =2x(x-3)+1(x-3)$
$\displaystyle =(x-3)(2x+1)$
$\displaystyle \text{The zeroes of }p(x)\text{ are given by }p(x)=0.$
$\displaystyle (x-3)(2x+1)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=-\frac{1}{2}$
$\displaystyle \text{Thus, }\alpha=3\text{ and }\beta=-\frac{1}{2}.$
$\displaystyle \text{The zeroes of the required polynomial are }\alpha-1=2\text{ and }\beta-1=-\frac{3}{2}.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle S=2-\frac{3}{2}=\frac{1}{2}$
$\displaystyle P=2\left(-\frac{3}{2}\right)=-3$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left(x^2-\frac{1}{2}x-3\right)$
$\displaystyle =\frac{k}{2}(2x^2-x-6)$
$\displaystyle \text{Taking }\frac{k}{2}=1,\text{ the required polynomial is}$
$\displaystyle 2x^2-x-6.$
$\\$

$\displaystyle \textbf{Question 18. }\text{Find the zeroes of the polynomial }p(x)=3x^2-4x-4.\text{ Hence, write a}$
$\displaystyle \text{polynomial whose each of the zeroes is 2 more than the zeroes of }p(x).\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle p(x)=3x^2-4x-4$
$\displaystyle =3x^2-6x+2x-4$
$\displaystyle =3x(x-2)+2(x-2)$
$\displaystyle =(x-2)(3x+2)$
$\displaystyle \text{The zeroes of }p(x)\text{ are given by }p(x)=0.$
$\displaystyle (x-2)(3x+2)=0$
$\displaystyle \Rightarrow x=2\text{ or }x=-\frac{2}{3}$
$\displaystyle \text{Thus, }\alpha=2\text{ and }\beta=-\frac{2}{3}.$
$\displaystyle \text{The zeroes of the required polynomial are }\alpha+2=4\text{ and }\beta+2=\frac{4}{3}.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle S=4+\frac{4}{3}=\frac{16}{3}$
$\displaystyle P=4\times\frac{4}{3}=\frac{16}{3}$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left(x^2-\frac{16}{3}x+\frac{16}{3}\right)$
$\displaystyle =\frac{k}{3}(3x^2-16x+16)$
$\displaystyle \text{Taking }\frac{k}{3}=1,\text{ the required polynomial is}$
$\displaystyle 3x^2-16x+16.$
$\\$

$\displaystyle \textbf{Question 19. }\text{Find the zeroes of the polynomial }q(x)=8x^2-2x-3.\text{ Hence, find a}$
$\displaystyle \text{polynomial whose zeroes are 2 less than the zeroes of }q(x).\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle q(x)=8x^2-2x-3$
$\displaystyle =8x^2+4x-6x-3$
$\displaystyle =4x(2x+1)-3(2x+1)$
$\displaystyle =(2x+1)(4x-3)$
$\displaystyle \text{The zeroes of }q(x)\text{ are given by }q(x)=0.$
$\displaystyle (2x+1)(4x-3)=0$
$\displaystyle \Rightarrow x=-\frac{1}{2}\text{ or }x=\frac{3}{4}$
$\displaystyle \text{Thus, }\alpha=-\frac{1}{2}\text{ and }\beta=\frac{3}{4}.$
$\displaystyle \text{The zeroes of the required polynomial are }\alpha-2=-\frac{5}{2}\text{ and }\beta-2=-\frac{5}{4}.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle S=-\frac{5}{2}-\frac{5}{4}=-\frac{15}{4}$
$\displaystyle P=\left(-\frac{5}{2}\right)\left(-\frac{5}{4}\right)=\frac{25}{8}$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k\left(x^2+\frac{15}{4}x+\frac{25}{8}\right)$
$\displaystyle =\frac{k}{8}(8x^2+30x+25)$
$\displaystyle \text{Taking }\frac{k}{8}=1,\text{ the required polynomial is}$
$\displaystyle 8x^2+30x+25.$
$\\$

$\displaystyle \textbf{Question 20. }\text{Find the zeroes of the polynomial }r(x)=4x^2+3x-1.\text{ Hence, write}$
$\displaystyle \text{a polynomial whose zeroes are reciprocal of the zeroes of polynomial }r(x).\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle r(x)=4x^2+3x-1$
$\displaystyle =4x^2+4x-x-1$
$\displaystyle =4x(x+1)-1(x+1)$
$\displaystyle =(x+1)(4x-1)$
$\displaystyle \text{The zeroes of }r(x)\text{ are given by }r(x)=0.$
$\displaystyle (x+1)(4x-1)=0$
$\displaystyle \Rightarrow x=-1\text{ or }x=\frac{1}{4}$
$\displaystyle \text{Thus, }\alpha=-1\text{ and }\beta=\frac{1}{4}.$
$\displaystyle \text{The zeroes of the required polynomial are }\frac{1}{\alpha}=-1\text{ and }\frac{1}{\beta}=4.$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeroes of the required polynomial.}$
$\displaystyle S=-1+4=3$
$\displaystyle P=(-1)(4)=-4$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k(x^2-3x-4)$
$\displaystyle \text{Taking }k=1,\text{ the required polynomial is}$
$\displaystyle x^2-3x-4.$
$\\$

$\displaystyle \textbf{Question 21. }\text{If }\alpha\text{ and }\beta\text{ are the zeros of the quadratic polynomial}$
$\displaystyle f(x)=x^2-1,\text{ find a quadratic polynomial }\text{whose zeros are }\frac{2\alpha}{\beta}\text{ and }\frac{2\beta}{\alpha}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeros of }f(x)=x^2-1.$
$\displaystyle \therefore \alpha+\beta=-\frac{0}{1}=0,\quad \alpha\beta=\frac{-1}{1}=-1$
$\displaystyle \text{Let }S\text{ and }P\text{ denote respectively the sum and product of zeros of the required polynomial.}$
$\displaystyle \text{Then,}$
$\displaystyle S=\frac{2\alpha}{\beta}+\frac{2\beta}{\alpha}$
$\displaystyle =2\left(\frac{\alpha^2+\beta^2}{\alpha\beta}\right)$
$\displaystyle =2\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)$
$\displaystyle =2\left(\frac{0^2-2(-1)}{-1}\right)$
$\displaystyle =2\left(\frac{2}{-1}\right)$
$\displaystyle =-4$
$\displaystyle \text{and,}$
$\displaystyle P=\frac{2\alpha}{\beta}\times\frac{2\beta}{\alpha}=4$
$\displaystyle \text{Hence, the required polynomial is given by}$
$\displaystyle g(x)=k(x^2-Sx+P)$
$\displaystyle =k(x^2+4x+4)$
$\displaystyle \text{Taking }k=1,\text{ the required polynomial is }x^2+4x+4.$
$\\$

$\displaystyle \textbf{Question 22. }\text{If the squared difference of the zeros of the quadratic polynomial }$
$\displaystyle f(x)=x^2+px+45\text{ is equal to }144,\text{ find the value of }p.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha\text{ and }\beta\text{ be the zeros of }f(x)=x^2+px+45.$
$\displaystyle \therefore \alpha+\beta=-\frac{p}{1}=-p,\quad \alpha\beta=\frac{45}{1}=45$
$\displaystyle \text{It is given that the squared difference of the zeros is }144.$
$\displaystyle \therefore (\alpha-\beta)^2=144$
$\displaystyle \Rightarrow (\alpha+\beta)^2-4\alpha\beta=144$
$\displaystyle \Rightarrow (-p)^2-4(45)=144$
$\displaystyle \Rightarrow p^2-180=144$
$\displaystyle \Rightarrow p^2=324$
$\displaystyle \Rightarrow p=\pm 18$
$\displaystyle \text{Hence, }p=18\text{ or }p=-18.$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }\alpha\text{ and }\beta\text{ are the zeros of the quadratic polynomial }$
$\displaystyle f(x)=x^2-px+q,\text{ prove that} \ \frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}=\frac{p^4}{q^2}-\frac{4p^2}{q}+2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeros of }f(x)=x^2-px+q.$
$\displaystyle \therefore \alpha+\beta=-\frac{-p}{1}=p,\quad \alpha\beta=\frac{q}{1}=q$
$\displaystyle \text{Now,}$
$\displaystyle \frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}=\frac{\alpha^4+\beta^4}{\alpha^2\beta^2}$
$\displaystyle =\frac{\alpha^4+\beta^4}{(\alpha\beta)^2}$
$\displaystyle \text{Also,}$
$\displaystyle \alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2$
$\displaystyle =\left\{(\alpha+\beta)^2-2\alpha\beta\right\}^2-2(\alpha\beta)^2$
$\displaystyle =(p^2-2q)^2-2q^2$
$\displaystyle =p^4-4p^2q+4q^2-2q^2$
$\displaystyle =p^4-4p^2q+2q^2$
$\displaystyle \therefore \frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}=\frac{p^4-4p^2q+2q^2}{q^2}$
$\displaystyle =\frac{p^4}{q^2}-\frac{4p^2q}{q^2}+\frac{2q^2}{q^2}$
$\displaystyle =\frac{p^4}{q^2}-\frac{4p^2}{q}+2$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 24. }\text{If }\alpha\text{ and }\beta\text{ are the zeros of the quadratic polynomial } \\ f(x)=ax^2+bx+c,\text{ then evaluate:}$
$\displaystyle \textbf{(i) }\frac{1}{a\alpha+b}+\frac{1}{a\beta+b}\qquad \textbf{(ii) }\frac{\beta}{a\alpha+b}+\frac{\alpha}{a\beta+b}$
$\displaystyle \textbf{(iii) }a\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)+b\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeros of }f(x)=ax^2+bx+c.$
$\displaystyle \therefore \alpha+\beta=-\frac{b}{a},\quad \alpha\beta=\frac{c}{a}$

$\displaystyle \textbf{(i)}$
$\displaystyle \frac{1}{a\alpha+b}+\frac{1}{a\beta+b}$
$\displaystyle =\frac{a\beta+b+a\alpha+b}{(a\alpha+b)(a\beta+b)}$
$\displaystyle =\frac{a(\alpha+\beta)+2b}{a^2\alpha\beta+ab(\alpha+\beta)+b^2}$
$\displaystyle =\frac{a\left(-\frac{b}{a}\right)+2b}{a^2\left(\frac{c}{a}\right)+ab\left(-\frac{b}{a}\right)+b^2}$
$\displaystyle =\frac{-b+2b}{ac-b^2+b^2}$
$\displaystyle =\frac{b}{ac}$

$\displaystyle \textbf{(ii)}$
$\displaystyle \frac{\beta}{a\alpha+b}+\frac{\alpha}{a\beta+b}$
$\displaystyle =\frac{\beta(a\beta+b)+\alpha(a\alpha+b)}{(a\alpha+b)(a\beta+b)}$
$\displaystyle =\frac{a(\alpha^2+\beta^2)+2b\alpha\beta}{ac}$
$\displaystyle =\frac{a\{(\alpha+\beta)^2-2\alpha\beta\}+2b\alpha\beta}{ac}$
$\displaystyle =\frac{a\left\{\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right\}+2b\left(\frac{c}{a}\right)}{ac}$
$\displaystyle =\frac{\frac{b^2}{a}-2c+\frac{2bc}{a}}{ac}$
$\displaystyle =\frac{b^2-2ac+2bc}{a^2c}$

$\displaystyle \textbf{(iii)}$
$\displaystyle a\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)+b\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)$
$\displaystyle =a\left(\frac{\alpha^3+\beta^3}{\alpha\beta}\right)+b\left(\frac{\alpha^2+\beta^2}{\alpha\beta}\right)$
$\displaystyle =a\left(\frac{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}{\alpha\beta}\right)+b\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)$
$\displaystyle =a\left(\frac{\left(-\frac{b}{a}\right)^3-3\left(\frac{c}{a}\right)\left(-\frac{b}{a}\right)}{\frac{c}{a}}\right)+b\left(\frac{\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)}{\frac{c}{a}}\right)$
$\displaystyle =a\left(\frac{3abc-b^3}{a^2c}\right)+b\left(\frac{b^2-2ac}{ac}\right)$
$\displaystyle =\frac{3abc-b^3}{ac}+\frac{b^3-2abc}{ac}$
$\displaystyle =\frac{abc}{ac}$
$\displaystyle =b$
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$\displaystyle \textbf{Question 25. }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial } \\ f(x)=x^2+x-2,\text{ find the value of}$
$\displaystyle \textbf{(i) }\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\qquad [\mathrm{CBSE}\ 2024]$
$\displaystyle \textbf{(ii) }\frac{1}{\alpha}-\frac{1}{\beta}\qquad [\mathrm{CBSE}\ 2024]$

$\displaystyle \textbf{(i)}\ \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }f(x)=x^2+x-2.$
$\displaystyle \therefore \alpha+\beta=-\frac{1}{1}=-1,\quad \alpha\beta=\frac{-2}{1}=-2$
$\displaystyle \text{Now,}$
$\displaystyle \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}$
$\displaystyle =\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
$\displaystyle =\frac{(-1)^2-2(-2)}{-2}$
$\displaystyle =\frac{1+4}{-2}$
$\displaystyle =-\frac{5}{2}$
$\displaystyle \text{Hence, }\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=-\frac{5}{2}.$
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$\displaystyle \textbf{(ii)}\ \text{Answer:}$
$\displaystyle \text{Given that }\alpha\text{ and }\beta\text{ are the zeroes of }f(x)=x^2+x-2.$
$\displaystyle \therefore \alpha+\beta=-1,\quad \alpha\beta=-2$
$\displaystyle \text{Also,}$
$\displaystyle f(x)=x^2+x-2$
$\displaystyle =x^2+2x-x-2$
$\displaystyle =x(x+2)-1(x+2)$
$\displaystyle =(x+2)(x-1)$
$\displaystyle \therefore \alpha=1,\quad \beta=-2$
$\displaystyle \text{Now,}$
$\displaystyle \frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}$
$\displaystyle =\frac{-2-1}{-2}$
$\displaystyle =\frac{3}{2}$
$\displaystyle \text{Hence, }\frac{1}{\alpha}-\frac{1}{\beta}=\frac{3}{2}.$
$\\$