Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Determine the nature of the roots of the following quadratic equations:}$
$\displaystyle \text{(i) }2x^2-3x+5=0$
$\displaystyle \text{(ii) }2x^2-6x+3=0$
$\displaystyle \text{(iii) }3x^2-4\sqrt{3}x+4=0$
$\displaystyle \text{(iv) }4x^2+4\sqrt{3}x+3=0\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }2x^2-3x+5=0$
$\displaystyle \text{Here, }a=2,\ b=-3,\ c=5.$
$\displaystyle D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31<0$
$\displaystyle \therefore \text{The equation has no real roots.}$

$\displaystyle \text{(ii) }2x^2-6x+3=0$
$\displaystyle \text{Here, }a=2,\ b=-6,\ c=3.$
$\displaystyle D=b^2-4ac=(-6)^2-4(2)(3)=36-24=12>0$
$\displaystyle \therefore \text{The equation has real and distinct roots.}$

$\displaystyle \text{(iii) }3x^2-4\sqrt{3}x+4=0$
$\displaystyle \text{Here, }a=3,\ b=-4\sqrt{3},\ c=4.$
$\displaystyle D=b^2-4ac=(-4\sqrt{3})^2-4(3)(4)=48-48=0$
$\displaystyle \therefore \text{The equation has real and equal roots.}$

$\displaystyle \text{(iv) }4x^2+4\sqrt{3}x+3=0$
$\displaystyle \text{Here, }a=4,\ b=4\sqrt{3},\ c=3.$
$\displaystyle D=b^2-4ac=(4\sqrt{3})^2-4(4)(3)=48-48=0$
$\displaystyle \therefore \text{The equation has real and equal roots.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the values of }k\text{ for which the roots are real and equal in each}$
$\displaystyle \text{of the following quadratic equations:}$
$\displaystyle \text{(i) }4x^2-2(k+1)x+(k+4)=0$
$\displaystyle \text{(ii) }4x^2-2(k+1)x+(k+1)=0\hfill\text{[CBSE 2017]}$
$\displaystyle \text{(iii) }x^2-2(k+1)x+k^2=0\hfill\text{[CBSE 2001C, 2013]}$
$\displaystyle \text{(iv) }k^2x^2-2(2k-1)x+4=0\hfill\text{[CBSE 2001 C]}$
$\displaystyle \text{(v) }(k+1)x^2-2(k-1)x+1=0\hfill\text{[CBSE 2002 C]}$
$\displaystyle \text{(vi) }x^2+k(2x+k-1)+2=0\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For real and equal roots, }D=b^2-4ac=0.$
$\displaystyle \text{(i) }4x^2-2(k+1)x+(k+4)=0$
$\displaystyle \text{Here, }a=4,\ b=-2(k+1),\ c=k+4.$
$\displaystyle D=\{-2(k+1)\}^2-4(4)(k+4)$
$\displaystyle =4(k+1)^2-16(k+4)$
$\displaystyle =4(k^2-2k-15)$
$\displaystyle =4(k-5)(k+3)$
$\displaystyle D=0\Rightarrow (k-5)(k+3)=0$
$\displaystyle \therefore k=5\text{ or }k=-3$

$\displaystyle \text{(ii) }4x^2-2(k+1)x+(k+1)=0$
$\displaystyle \text{Here, }a=4,\ b=-2(k+1),\ c=k+1.$
$\displaystyle D=\{-2(k+1)\}^2-4(4)(k+1)$
$\displaystyle =4(k+1)^2-16(k+1)$
$\displaystyle =4(k+1)(k-3)$
$\displaystyle D=0\Rightarrow (k+1)(k-3)=0$
$\displaystyle \therefore k=-1\text{ or }k=3$

$\displaystyle \text{(iii) }x^2-2(k+1)x+k^2=0$
$\displaystyle \text{Here, }a=1,\ b=-2(k+1),\ c=k^2.$
$\displaystyle D=\{-2(k+1)\}^2-4(1)(k^2)$
$\displaystyle =4\{(k+1)^2-k^2\}=4(2k+1)$
$\displaystyle D=0\Rightarrow 2k+1=0$
$\displaystyle \therefore k=-\frac{1}{2}$

$\displaystyle \text{(iv) }k^2x^2-2(2k-1)x+4=0$
$\displaystyle \text{Here, }a=k^2,\ b=-2(2k-1),\ c=4.$
$\displaystyle D=\{-2(2k-1)\}^2-4(k^2)(4)$
$\displaystyle =4(2k-1)^2-16k^2$
$\displaystyle =4(1-4k)$
$\displaystyle D=0\Rightarrow 1-4k=0$
$\displaystyle \therefore k=\frac{1}{4}$

$\displaystyle \text{(v) }(k+1)x^2-2(k-1)x+1=0$
$\displaystyle \text{Here, }a=k+1,\ b=-2(k-1),\ c=1.$
$\displaystyle D=\{-2(k-1)\}^2-4(k+1)(1)$
$\displaystyle =4\{(k-1)^2-(k+1)\}$
$\displaystyle =4(k^2-3k)=4k(k-3)$
$\displaystyle D=0\Rightarrow k(k-3)=0$
$\displaystyle \therefore k=0\text{ or }k=3$

$\displaystyle \text{(vi) }x^2+k(2x+k-1)+2=0$
$\displaystyle \Rightarrow x^2+2kx+k^2-k+2=0$
$\displaystyle \text{Here, }a=1,\ b=2k,\ c=k^2-k+2.$
$\displaystyle D=(2k)^2-4(1)(k^2-k+2)$
$\displaystyle =4k^2-4k^2+4k-8=4(k-2)$
$\displaystyle D=0\Rightarrow k-2=0$
$\displaystyle \therefore k=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In each of the following, determine the values of }k\text{ for which the given}$
$\displaystyle \text{quadratic equation has equal roots:}$
$\displaystyle \text{(i) }2x^2+kx+3=0$
$\displaystyle \text{(ii) }kx(x-2)+6=0\hfill\text{[2013]}$
$\displaystyle \text{(iii) }x^2-4kx+k=0\hfill\text{[CBSE 2012]}$
$\displaystyle \text{(iv) }kx(x-2\sqrt{5})+10=0\hfill\text{[CBSE 2013]}$
$\displaystyle \text{(v) }kx(x-3)+9=0\hfill\text{[CBSE 2014]}$
$\displaystyle \text{(vi) }4x^2+kx+3=0\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For equal roots, }D=b^2-4ac=0.$
$\displaystyle \text{(i) }2x^2+kx+3=0$
$\displaystyle \text{Here, }a=2,\ b=k,\ c=3.$
$\displaystyle D=k^2-4(2)(3)=k^2-24$
$\displaystyle D=0\Rightarrow k^2-24=0$
$\displaystyle \Rightarrow k^2=24$
$\displaystyle \therefore k=\pm2\sqrt{6}$

$\displaystyle \text{(ii) }kx(x-2)+6=0$
$\displaystyle \Rightarrow kx^2-2kx+6=0$
$\displaystyle \text{Here, }a=k,\ b=-2k,\ c=6,\quad k\ne0.$
$\displaystyle D=(-2k)^2-4(k)(6)=4k^2-24k$
$\displaystyle =4k(k-6)$
$\displaystyle D=0\Rightarrow k=0\text{ or }k=6$
$\displaystyle \text{But }k\ne0\text{ since the equation must be quadratic.}$
$\displaystyle \therefore k=6$

$\displaystyle \text{(iii) }x^2-4kx+k=0$
$\displaystyle \text{Here, }a=1,\ b=-4k,\ c=k.$
$\displaystyle D=(-4k)^2-4(1)(k)=16k^2-4k$
$\displaystyle =4k(4k-1)$
$\displaystyle D=0\Rightarrow k=0\text{ or }4k-1=0$
$\displaystyle \therefore k=0\text{ or }k=\frac{1}{4}$

$\displaystyle \text{(iv) }kx(x-2\sqrt{5})+10=0$
$\displaystyle \Rightarrow kx^2-2\sqrt{5}kx+10=0$
$\displaystyle \text{Here, }a=k,\ b=-2\sqrt{5}k,\ c=10,\quad k\ne0.$
$\displaystyle D=(-2\sqrt{5}k)^2-4(k)(10)=20k^2-40k$
$\displaystyle =20k(k-2)$
$\displaystyle D=0\Rightarrow k=0\text{ or }k=2$
$\displaystyle \text{But }k\ne0\text{ since the equation must be quadratic.}$
$\displaystyle \therefore k=2$

$\displaystyle \text{(v) }kx(x-3)+9=0$
$\displaystyle \Rightarrow kx^2-3kx+9=0$
$\displaystyle \text{Here, }a=k,\ b=-3k,\ c=9,\quad k\ne0.$
$\displaystyle D=(-3k)^2-4(k)(9)=9k^2-36k$
$\displaystyle =9k(k-4)$
$\displaystyle D=0\Rightarrow k=0\text{ or }k=4$
$\displaystyle \text{But }k\ne0\text{ since the equation must be quadratic.}$
$\displaystyle \therefore k=4$

$\displaystyle \text{(vi) }4x^2+kx+3=0$
$\displaystyle \text{Here, }a=4,\ b=k,\ c=3.$
$\displaystyle D=k^2-4(4)(3)=k^2-48$
$\displaystyle D=0\Rightarrow k^2-48=0$
$\displaystyle \Rightarrow k^2=48$
$\displaystyle \therefore k=\pm4\sqrt{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the values of }k\text{ for which the given quadratic equation has real}$
$\displaystyle \text{and distinct roots:}$
$\displaystyle \text{(i) }kx^2+2x+1=0$
$\displaystyle \text{(ii) }kx^2+6x+1=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For real and distinct roots, }D=b^2-4ac>0.$
$\displaystyle \text{(i) }kx^2+2x+1=0$
$\displaystyle \text{Here, }a=k,\ b=2,\ c=1,\quad k\ne0.$
$\displaystyle D=2^2-4(k)(1)=4-4k=4(1-k)$
$\displaystyle D>0\Rightarrow 4(1-k)>0$
$\displaystyle \Rightarrow 1-k>0$
$\displaystyle \Rightarrow k<1$
$\displaystyle \therefore k<1,\quad k\ne0$

$\displaystyle \text{(ii) }kx^2+6x+1=0$
$\displaystyle \text{Here, }a=k,\ b=6,\ c=1,\quad k\ne0.$
$\displaystyle D=6^2-4(k)(1)=36-4k=4(9-k)$
$\displaystyle D>0\Rightarrow 4(9-k)>0$
$\displaystyle \Rightarrow 9-k>0$
$\displaystyle \Rightarrow k<9$
$\displaystyle \therefore k<9,\quad k\ne0$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{For what values of }k,\ (4-k)x^2+(2k+4)x+(8k+1)$
$\displaystyle \text{is a perfect square?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given expression to be a perfect square, its discriminant must be zero.}$
$\displaystyle D=(2k+4)^2-4(4-k)(8k+1)$
$\displaystyle =4(k+2)^2-4(4-k)(8k+1)$
$\displaystyle =4\{k^2+4k+4-(31k+4-8k^2)\}$
$\displaystyle =4(9k^2-27k)$
$\displaystyle =36k(k-3)$
$\displaystyle D=0\Rightarrow 36k(k-3)=0$
$\displaystyle \therefore k=0\text{ or }k=3$
$\displaystyle \text{For }k=0,\quad 4x^2+4x+1=(2x+1)^2.$
$\displaystyle \text{For }k=3,\quad x^2+10x+25=(x+5)^2.$
$\displaystyle \therefore \text{The required values are }k=0\text{ and }k=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{(i) Find the values of }k\text{ for which the quadratic equation}$
$\displaystyle (3k+1)x^2+2(k+1)x+1=0\text{ has equal roots. Also, find the roots.}$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{(ii) Write all the values of }k\text{ for which the quadratic equation }x^2+kx+16=0$
$\displaystyle \text{has equal roots. Find the roots of the equation so obtained.}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{(iii) Find the value of }p\text{ for which the equation }px(x-2)+6=0\text{ has two equal}$
$\displaystyle \text{roots. Also, find the roots.}\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For equal roots, }D=b^2-4ac=0.$
$\displaystyle \text{(i) }(3k+1)x^2+2(k+1)x+1=0$
$\displaystyle \text{Here, }a=3k+1,\ b=2(k+1),\ c=1.$
$\displaystyle D=\{2(k+1)\}^2-4(3k+1)(1)$
$\displaystyle =4\{(k+1)^2-(3k+1)\}$
$\displaystyle =4(k^2-k)=4k(k-1)$
$\displaystyle D=0\Rightarrow k(k-1)=0$
$\displaystyle \therefore k=0\text{ or }k=1$
$\displaystyle \text{For }k=0,\text{ the equation becomes }x^2+2x+1=0.$
$\displaystyle \Rightarrow (x+1)^2=0$
$\displaystyle \therefore \text{Both roots are }x=-1.$
$\displaystyle \text{For }k=1,\text{ the equation becomes }4x^2+4x+1=0.$
$\displaystyle \Rightarrow (2x+1)^2=0$
$\displaystyle \therefore \text{Both roots are }x=-\frac{1}{2}.$

$\displaystyle \text{(ii) }x^2+kx+16=0$
$\displaystyle \text{Here, }a=1,\ b=k,\ c=16.$
$\displaystyle D=k^2-4(1)(16)=k^2-64$
$\displaystyle D=0\Rightarrow k^2-64=0$
$\displaystyle \Rightarrow (k-8)(k+8)=0$
$\displaystyle \therefore k=8\text{ or }k=-8$
$\displaystyle \text{For }k=8,\text{ the equation becomes }x^2+8x+16=0.$
$\displaystyle \Rightarrow (x+4)^2=0$
$\displaystyle \therefore \text{Both roots are }x=-4.$
$\displaystyle \text{For }k=-8,\text{ the equation becomes }x^2-8x+16=0.$
$\displaystyle \Rightarrow (x-4)^2=0$
$\displaystyle \therefore \text{Both roots are }x=4.$

$\displaystyle \text{(iii) }px(x-2)+6=0$
$\displaystyle \Rightarrow px^2-2px+6=0$
$\displaystyle \text{Here, }a=p,\ b=-2p,\ c=6,\quad p\ne0.$
$\displaystyle D=(-2p)^2-4(p)(6)=4p^2-24p$
$\displaystyle =4p(p-6)$
$\displaystyle D=0\Rightarrow p=0\text{ or }p=6$
$\displaystyle \text{But }p\ne0\text{ since the equation must be quadratic.}$
$\displaystyle \therefore p=6$
$\displaystyle \text{Substituting }p=6,\text{ we get }6x^2-12x+6=0.$
$\displaystyle \Rightarrow x^2-2x+1=0$
$\displaystyle \Rightarrow (x-1)^2=0$
$\displaystyle \therefore \text{Both roots are }x=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the values of }p\text{ for which the quadratic equation}$
$\displaystyle (2p+1)x^2-(7p+2)x+(7p-3)=0\text{ has equal roots. Also, find these roots.}$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For equal roots, }D=b^2-4ac=0.$
$\displaystyle \text{Here, }a=2p+1,\ b=-(7p+2),\ c=7p-3.$
$\displaystyle D=(7p+2)^2-4(2p+1)(7p-3)$
$\displaystyle =49p^2+28p+4-4(14p^2+p-3)$
$\displaystyle =49p^2+28p+4-56p^2-4p+12$
$\displaystyle =-7p^2+24p+16$
$\displaystyle D=0\Rightarrow -7p^2+24p+16=0$
$\displaystyle \Rightarrow 7p^2-24p-16=0$
$\displaystyle \Rightarrow 7p^2-28p+4p-16=0$
$\displaystyle \Rightarrow 7p(p-4)+4(p-4)=0$
$\displaystyle \Rightarrow (p-4)(7p+4)=0$
$\displaystyle \therefore p=4\text{ or }p=-\frac{4}{7}$
$\displaystyle \text{For }p=4,\text{ the equation becomes }9x^2-30x+25=0.$
$\displaystyle \Rightarrow (3x-5)^2=0$
$\displaystyle \therefore \text{Both roots are }x=\frac{5}{3}.$
$\displaystyle \text{For }p=-\frac{4}{7},\text{ the equation becomes }-\frac{1}{7}x^2+2x-7=0.$
$\displaystyle \Rightarrow x^2-14x+49=0$
$\displaystyle \Rightarrow (x-7)^2=0$
$\displaystyle \therefore \text{Both roots are }x=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{(i) Find the value of }p\text{ for which the quadratic equation}$
$\displaystyle (p+1)x^2-6(p+1)x+3(p+9)=0,\quad p\ne-1\text{ has equal roots. Hence, find}$
$\displaystyle \text{the roots of the equation.}\hfill\text{[CBSE 2015, 2024]}$
$\displaystyle \text{(ii) Find the value(s) of }p\text{ for which the quadratic equation given as}$
$\displaystyle (p+4)x^2-(p+1)x+1=0\text{ has real and equal roots. Also, find the roots of}$
$\displaystyle \text{the equation(s) so obtained.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{(iii) Find the smallest value of }p\text{ for which the quadratic equation}$
$\displaystyle x^2-2(p+1)x+p^2=0\text{ has real roots. Hence, find the roots of the equation}$
$\displaystyle \text{so obtained.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(p+1)x^2-6(p+1)x+3(p+9)=0$
$\displaystyle \text{Here, }a=p+1,\ b=-6(p+1),\ c=3(p+9).$
$\displaystyle \text{For equal roots, }D=b^2-4ac=0.$
$\displaystyle \{-6(p+1)\}^2-4(p+1)\{3(p+9)\}=0$
$\displaystyle \Rightarrow 36(p+1)^2-12(p+1)(p+9)=0$
$\displaystyle \Rightarrow 12(p+1)\{3(p+1)-(p+9)\}=0$
$\displaystyle \Rightarrow 24(p+1)(p-3)=0$
$\displaystyle \Rightarrow p=-1\text{ or }p=3$
$\displaystyle \text{But }p\ne-1.$
$\displaystyle \therefore p=3$
$\displaystyle \text{Substituting }p=3,\text{ we get }4x^2-24x+36=0.$
$\displaystyle \Rightarrow x^2-6x+9=0$
$\displaystyle \Rightarrow (x-3)^2=0$
$\displaystyle \therefore \text{Both roots are }x=3.$

$\displaystyle \text{(ii) }(p+4)x^2-(p+1)x+1=0$
$\displaystyle \text{Here, }a=p+4,\ b=-(p+1),\ c=1.$
$\displaystyle \text{For real and equal roots, }D=b^2-4ac=0.$
$\displaystyle (p+1)^2-4(p+4)=0$
$\displaystyle \Rightarrow p^2+2p+1-4p-16=0$
$\displaystyle \Rightarrow p^2-2p-15=0$
$\displaystyle \Rightarrow (p-5)(p+3)=0$
$\displaystyle \therefore p=5\text{ or }p=-3$
$\displaystyle \text{For }p=5,\text{ the equation becomes }9x^2-6x+1=0.$
$\displaystyle \Rightarrow (3x-1)^2=0$
$\displaystyle \therefore \text{Both roots are }x=\frac{1}{3}.$
$\displaystyle \text{For }p=-3,\text{ the equation becomes }x^2+2x+1=0.$
$\displaystyle \Rightarrow (x+1)^2=0$
$\displaystyle \therefore \text{Both roots are }x=-1.$

$\displaystyle \text{(iii) }x^2-2(p+1)x+p^2=0$
$\displaystyle \text{Here, }a=1,\ b=-2(p+1),\ c=p^2.$
$\displaystyle \text{For real roots, }D=b^2-4ac\ge0.$
$\displaystyle \{-2(p+1)\}^2-4(1)(p^2)\ge0$
$\displaystyle \Rightarrow 4(p+1)^2-4p^2\ge0$
$\displaystyle \Rightarrow 4(2p+1)\ge0$
$\displaystyle \Rightarrow 2p+1\ge0$
$\displaystyle \Rightarrow p\ge-\frac{1}{2}$
$\displaystyle \therefore \text{The smallest value of }p=-\frac{1}{2}.$
$\displaystyle \text{Substituting }p=-\frac{1}{2},\text{ we get }x^2-x+\frac{1}{4}=0.$
$\displaystyle \Rightarrow \left(x-\frac{1}{2}\right)^2=0$
$\displaystyle \therefore \text{Both roots are }x=\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the least positive value of }k\text{ for which the equation }x^2+kx+4=0$
$\displaystyle \text{has real roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }x^2+kx+4=0.$
$\displaystyle \text{Here, }a=1,\ b=k,\ c=4.$
$\displaystyle \text{For real roots, }D=b^2-4ac\ge0.$
$\displaystyle \Rightarrow k^2-4(1)(4)\ge0$
$\displaystyle \Rightarrow k^2-16\ge0$
$\displaystyle \Rightarrow (k-4)(k+4)\ge0$
$\displaystyle \Rightarrow k\le-4\text{ or }k\ge4$
$\displaystyle \therefore \text{The least positive value of }k\text{ is }4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If the roots of the equation }(b-c)x^2+(c-a)x+(a-b)=0$
$\displaystyle \text{are equal, then prove that }2b=a+c.\hfill\text{[CBSE 2002 C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }(b-c)x^2+(c-a)x+(a-b)=0.$
$\displaystyle \text{Here, }A=b-c,\ B=c-a,\ C=a-b.$
$\displaystyle \text{Since the roots are equal, }D=B^2-4AC=0.$
$\displaystyle \Rightarrow (c-a)^2-4(b-c)(a-b)=0$
$\displaystyle \Rightarrow a^2-2ac+c^2-4(ab-b^2-ac+bc)=0$
$\displaystyle \Rightarrow a^2+2ac+c^2-4ab-4bc+4b^2=0$
$\displaystyle \Rightarrow (a+c-2b)^2=0$
$\displaystyle \Rightarrow a+c-2b=0$
$\displaystyle \therefore 2b=a+c.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If the roots of the equation }(a^2+b^2)x^2-2(ac+bd)x$
$\displaystyle +(c^2+d^2)=0\text{ are equal, prove that }\frac{a}{b}=\frac{c}{d}.\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0.$
$\displaystyle \text{Here, }A=a^2+b^2,\ B=-2(ac+bd),\ C=c^2+d^2.$
$\displaystyle \text{Since the roots are equal, }D=B^2-4AC=0.$
$\displaystyle \Rightarrow \{-2(ac+bd)\}^2-4(a^2+b^2)(c^2+d^2)=0$
$\displaystyle \Rightarrow (ac+bd)^2-(a^2+b^2)(c^2+d^2)=0$
$\displaystyle \Rightarrow a^2c^2+2abcd+b^2d^2$
$\displaystyle \qquad -(a^2c^2+a^2d^2+b^2c^2+b^2d^2)=0$
$\displaystyle \Rightarrow 2abcd-a^2d^2-b^2c^2=0$
$\displaystyle \Rightarrow a^2d^2-2abcd+b^2c^2=0$
$\displaystyle \Rightarrow (ad-bc)^2=0$
$\displaystyle \Rightarrow ad=bc$
$\displaystyle \therefore \frac{a}{b}=\frac{c}{d},\quad b\ne0,\ d\ne0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If the equation }(1+m^2)x^2+2mcx+(c^2-a^2)=0\text{ has equal}$
$\displaystyle \text{roots, prove that }c^2=a^2(1+m^2).\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }(1+m^2)x^2+2mcx+(c^2-a^2)=0.$
$\displaystyle \text{Here, }A=1+m^2,\ B=2mc,\ C=c^2-a^2.$
$\displaystyle \text{Since the roots are equal, }D=B^2-4AC=0.$
$\displaystyle \Rightarrow (2mc)^2-4(1+m^2)(c^2-a^2)=0$
$\displaystyle \Rightarrow m^2c^2-(1+m^2)(c^2-a^2)=0$
$\displaystyle \Rightarrow m^2c^2-(1+m^2)c^2+(1+m^2)a^2=0$
$\displaystyle \Rightarrow -c^2+a^2(1+m^2)=0$
$\displaystyle \therefore c^2=a^2(1+m^2).$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Determine the nature of the roots of the following quadratic equations:}$
$\displaystyle \text{(i) }(x-2a)(x-2b)=4ab$
$\displaystyle \text{(ii) }9a^2b^2x^2-24abcdx+16c^2d^2=0,\quad a\ne0,\ b\ne0$
$\displaystyle \text{(iii) }2(a^2+b^2)x^2+2(a+b)x+1=0$
$\displaystyle \text{(iv) }(b+c)x^2-(a+b+c)x+a=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x-2a)(x-2b)=4ab$
$\displaystyle \Rightarrow x^2-2(a+b)x+4ab=4ab$
$\displaystyle \Rightarrow x^2-2(a+b)x=0$
$\displaystyle \text{Here, }A=1,\ B=-2(a+b),\ C=0.$
$\displaystyle D=B^2-4AC=\{-2(a+b)\}^2=4(a+b)^2\ge0$
$\displaystyle \therefore \text{The roots are real and distinct if }a+b\ne0,$
$\displaystyle \text{and real and equal if }a+b=0.$

$\displaystyle \text{(ii) }9a^2b^2x^2-24abcdx+16c^2d^2=0$
$\displaystyle \text{Here, }A=9a^2b^2,\ B=-24abcd,\ C=16c^2d^2.$
$\displaystyle D=B^2-4AC$
$\displaystyle =(-24abcd)^2-4(9a^2b^2)(16c^2d^2)$
$\displaystyle =576a^2b^2c^2d^2-576a^2b^2c^2d^2=0$
$\displaystyle \therefore \text{The roots are real and equal.}$

$\displaystyle \text{(iii) }2(a^2+b^2)x^2+2(a+b)x+1=0$
$\displaystyle \text{Here, }A=2(a^2+b^2),\ B=2(a+b),\ C=1.$
$\displaystyle D=B^2-4AC$
$\displaystyle =\{2(a+b)\}^2-4\{2(a^2+b^2)\}(1)$
$\displaystyle =4(a+b)^2-8(a^2+b^2)$
$\displaystyle =-4(a-b)^2\le0$
$\displaystyle \therefore \text{The roots are real and equal if }a=b,$
$\displaystyle \text{and there are no real roots if }a\ne b.$

$\displaystyle \text{(iv) }(b+c)x^2-(a+b+c)x+a=0$
$\displaystyle \text{Here, }A=b+c,\ B=-(a+b+c),\ C=a.$
$\displaystyle D=B^2-4AC$
$\displaystyle =(a+b+c)^2-4a(b+c)$
$\displaystyle =a^2+(b+c)^2-2a(b+c)$
$\displaystyle =\{a-(b+c)\}^2=(a-b-c)^2\ge0$
$\displaystyle \therefore \text{The roots are real and distinct if }a\ne b+c,$
$\displaystyle \text{and real and equal if }a=b+c.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If the roots of the equations }ax^2+2bx+c=0\text{ and}$
$\displaystyle bx^2-2\sqrt{ac}\,x+b=0\text{ are simultaneously real, then prove that }b^2=ac.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }ax^2+2bx+c=0,\text{ the roots are real.}$
$\displaystyle \therefore D_1=(2b)^2-4ac\ge0$
$\displaystyle \Rightarrow 4b^2-4ac\ge0$
$\displaystyle \Rightarrow b^2-ac\ge0$
$\displaystyle \Rightarrow b^2\ge ac$
$\displaystyle \text{For }bx^2-2\sqrt{ac}\,x+b=0,\text{ the roots are real.}$
$\displaystyle \therefore D_2=(-2\sqrt{ac})^2-4b^2\ge0$
$\displaystyle \Rightarrow 4ac-4b^2\ge0$
$\displaystyle \Rightarrow ac-b^2\ge0$
$\displaystyle \Rightarrow ac\ge b^2$
$\displaystyle \therefore b^2\ge ac\text{ and }ac\ge b^2.$
$\displaystyle \therefore b^2=ac.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If }p,q\text{ are real and }p\ne q,\text{ then show that the roots of the}$
$\displaystyle \text{equation }(p-q)x^2+5(p+q)x-2(p-q)=0\text{ are real and unequal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }(p-q)x^2+5(p+q)x-2(p-q)=0.$
$\displaystyle \text{Here, }a=p-q,\ b=5(p+q),\ c=-2(p-q).$
$\displaystyle D=b^2-4ac$
$\displaystyle =\{5(p+q)\}^2-4(p-q)\{-2(p-q)\}$
$\displaystyle =25(p+q)^2+8(p-q)^2$
$\displaystyle \text{Since }p\ne q,\quad (p-q)^2>0.$
$\displaystyle \therefore 25(p+q)^2+8(p-q)^2>0$
$\displaystyle \Rightarrow D>0$
$\displaystyle \therefore \text{The roots of the equation are real and unequal.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If the roots of the equation }(c^2-ab)x^2-2(a^2-bc)x$
$\displaystyle +(b^2-ac)=0\text{ are equal, prove that either }a=0\text{ or }a^3+b^3+c^3=3abc.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0.$
$\displaystyle \text{Since the roots are equal, }D=0.$
$\displaystyle \{-2(a^2-bc)\}^2-4(c^2-ab)(b^2-ac)=0$
$\displaystyle \Rightarrow (a^2-bc)^2-(c^2-ab)(b^2-ac)=0$
$\displaystyle \Rightarrow a^4-2a^2bc+b^2c^2$
$\displaystyle \qquad -(b^2c^2-ac^3-ab^3+a^2bc)=0$
$\displaystyle \Rightarrow a^4-3a^2bc+ac^3+ab^3=0$
$\displaystyle \Rightarrow a(a^3+b^3+c^3-3abc)=0$
$\displaystyle \therefore a=0\text{ or }a^3+b^3+c^3-3abc=0$
$\displaystyle \therefore a=0\text{ or }a^3+b^3+c^3=3abc.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Show that the equation }2(a^2+b^2)x^2+2(a+b)x+1=0$
$\displaystyle \text{has no real roots, when }a\ne b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }2(a^2+b^2)x^2+2(a+b)x+1=0.$
$\displaystyle \text{Here, }A=2(a^2+b^2),\ B=2(a+b),\ C=1.$
$\displaystyle D=B^2-4AC$
$\displaystyle =\{2(a+b)\}^2-4\{2(a^2+b^2)\}(1)$
$\displaystyle =4(a+b)^2-8(a^2+b^2)$
$\displaystyle =4(a^2+2ab+b^2-2a^2-2b^2)$
$\displaystyle =-4(a^2-2ab+b^2)$
$\displaystyle =-4(a-b)^2$
$\displaystyle \text{Since }a\ne b,\quad (a-b)^2>0.$
$\displaystyle \therefore D=-4(a-b)^2<0.$
$\displaystyle \therefore \text{The given equation has no real roots.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Prove that both the roots of the equation}$
$\displaystyle (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\text{ are real, but they are}$
$\displaystyle \text{equal only when }a=b=c.$
$\displaystyle \text{Answer:}$
$\displaystyle (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\displaystyle \Rightarrow x^2-(a+b)x+ab+x^2-(b+c)x+bc$
$\displaystyle \qquad +x^2-(c+a)x+ca=0$
$\displaystyle \Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
$\displaystyle \text{Here, }A=3,\ B=-2(a+b+c),\ C=ab+bc+ca.$
$\displaystyle D=B^2-4AC$
$\displaystyle =\{-2(a+b+c)\}^2-12(ab+bc+ca)$
$\displaystyle =4\{a^2+b^2+c^2-ab-bc-ca\}$
$\displaystyle =2\{(a-b)^2+(b-c)^2+(c-a)^2\}$
$\displaystyle \text{Since }(a-b)^2,(b-c)^2\text{ and }(c-a)^2\ge0,$
$\displaystyle D\ge0.$
$\displaystyle \therefore \text{Both the roots of the equation are real.}$
$\displaystyle \text{The roots are equal only when }D=0.$
$\displaystyle \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\displaystyle \Rightarrow a-b=0,\quad b-c=0,\quad c-a=0$
$\displaystyle \therefore a=b=c.$
$\displaystyle \therefore \text{Both roots are real and are equal only when }a=b=c.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If }a,b,c\text{ are real numbers such that }ac\ne0,\text{ then show that at least}$
$\displaystyle \text{one of the equations }ax^2+bx+c=0\text{ and }-ax^2+bx+c=0\text{ has real roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D_1\text{ and }D_2\text{ be the discriminants of the two equations respectively.}$
$\displaystyle \text{For }ax^2+bx+c=0,$
$\displaystyle D_1=b^2-4ac.$
$\displaystyle \text{For }-ax^2+bx+c=0,$
$\displaystyle D_2=b^2-4(-a)c=b^2+4ac.$
$\displaystyle \therefore D_1+D_2=(b^2-4ac)+(b^2+4ac)=2b^2\ge0.$
$\displaystyle \text{Hence, }D_1\text{ and }D_2\text{ cannot both be negative.}$
$\displaystyle \therefore \text{At least one of }D_1\text{ or }D_2\text{ is greater than or equal to zero.}$
$\displaystyle \therefore \text{At least one of the given quadratic equations has real roots.}$
$\displaystyle \\$