Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Which of the following are quadratic equations?}$
$\displaystyle \text{(i) }16x^2-3=(2x+5)(5x-3)$
$\displaystyle \text{(ii) }\sqrt{3x^2}-2x+\frac{1}{2}=0$
$\displaystyle \text{(iii) }x^2+\frac{1}{x^2}=5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }16x^2-3=(2x+5)(5x-3)$
$\displaystyle \Rightarrow 16x^2-3=10x^2+19x-15$
$\displaystyle \Rightarrow 6x^2-19x+12=0$
$\displaystyle \therefore \text{It is a quadratic equation.}$

$\displaystyle \text{(ii) }\sqrt{3}x^2-2x+\frac{1}{2}=0$
$\displaystyle \text{Clearly, }\sqrt{3}x^2-2x+\frac{1}{2}\text{ is a quadratic polynomial.}$
$\displaystyle \therefore \text{The given equation is a quadratic equation.}$

$\displaystyle \text{(iii) }x^2+\frac{1}{x^2}=5$
$\displaystyle \Rightarrow x^4+1=5x^2$
$\displaystyle \Rightarrow x^4-5x^2+1=0$
$\displaystyle \therefore \text{It is not a quadratic equation.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Which of the following are quadratic equations?}$
$\displaystyle \text{(iv) }x-\frac{3}{x}=x^2$
$\displaystyle \text{(v) }2x^2-\sqrt{3x}+9=0$
$\displaystyle \text{(vi) }x^2-2x-\sqrt{x}-5=0$
$\displaystyle \text{(vii) }(x+2)^3=x^3-4$
$\displaystyle \text{(viii) }x+\frac{1}{x}=1$
$\displaystyle \text{(ix) }x+\frac{1}{x}=x^2,\ x\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv) }x-\frac{3}{x}=x^2$
$\displaystyle \Rightarrow x^2-3=x^3$
$\displaystyle \Rightarrow x^3-x^2+3=0$
$\displaystyle \therefore \text{It is not a quadratic equation.}$

$\displaystyle \text{(v) }2x^2-\sqrt{3x}+9=0$
$\displaystyle \text{The equation contains the term }\sqrt{3x}=(3x)^{1/2}.$
$\displaystyle \therefore \text{It is not a quadratic equation.}$

$\displaystyle \text{(vi) }x^2-2x-\sqrt{x}-5=0$
$\displaystyle \text{The equation contains the term }\sqrt{x}=x^{1/2}.$
$\displaystyle \therefore \text{It is not a quadratic equation.}$

$\displaystyle \text{(vii) }(x+2)^3=x^3-4$
$\displaystyle \Rightarrow x^3+6x^2+12x+8=x^3-4$
$\displaystyle \Rightarrow 6x^2+12x+12=0$
$\displaystyle \Rightarrow x^2+2x+2=0$
$\displaystyle \therefore \text{It is a quadratic equation.}$

$\displaystyle \text{(viii) }x+\frac{1}{x}=1$
$\displaystyle \Rightarrow x^2+1=x$
$\displaystyle \Rightarrow x^2-x+1=0$
$\displaystyle \therefore \text{It is a quadratic equation.}$

$\displaystyle \text{(ix) }x+\frac{1}{x}=x^2,\ x\ne0$
$\displaystyle \Rightarrow x^2+1=x^3$
$\displaystyle \Rightarrow x^3-x^2-1=0$
$\displaystyle \therefore \text{It is not a quadratic equation.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Which of the following are quadratic equations?}$
$\displaystyle \text{(x) }\left(x+\frac{1}{x}\right)^2=3\left(x+\frac{1}{x}\right)+4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(x) }\left(x+\frac{1}{x}\right)^2=3\left(x+\frac{1}{x}\right)+4$
$\displaystyle \Rightarrow x^2+2+\frac{1}{x^2}=3x+\frac{3}{x}+4$
$\displaystyle \text{Multiplying both sides by }x^2,$
$\displaystyle x^4+2x^2+1=3x^3+3x+4x^2$
$\displaystyle \Rightarrow x^4-3x^3-2x^2-3x+1=0$
$\displaystyle \therefore \text{It is not a quadratic equation.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In each of the following, determine whether the given values are}$
$\displaystyle \text{solutions of the given equation or not:}$
$\displaystyle \text{(i) }x^2-3\sqrt{3}x+6=0,\quad x=\sqrt{3},\ x=-2\sqrt{3}$
$\displaystyle \text{(ii) }x+\frac{1}{x}=\frac{13}{6},\quad x=\frac{5}{6},\ x=\frac{4}{3}$
$\displaystyle \text{(iii) }2x^2-x+9=x^2+4x+3,\quad x=2,\ x=3$
$\displaystyle \text{(iv) }x^2-\sqrt{2}x-4=0,\quad x=-\sqrt{2},\ x=-2\sqrt{2}$
$\displaystyle \text{(v) }a^2x^2-3abx+2b^2=0,\quad x=\frac{a}{b},\ x=\frac{b}{a}$
$\displaystyle \text{(vi) }\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+16},\quad x=3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) For }x=\sqrt{3},$
$\displaystyle \text{LHS}=(\sqrt{3})^2-3\sqrt{3}(\sqrt{3})+6=3-9+6=0=\text{RHS}$
$\displaystyle \therefore x=\sqrt{3}\text{ is a solution.}$
$\displaystyle \text{For }x=-2\sqrt{3},$
$\displaystyle \text{LHS}=(-2\sqrt{3})^2-3\sqrt{3}(-2\sqrt{3})+6$
$\displaystyle =12+18+6=36\ne0=\text{RHS}$
$\displaystyle \therefore x=-2\sqrt{3}\text{ is not a solution.}$

$\displaystyle \text{(ii) For }x=\frac{5}{6},$
$\displaystyle \text{LHS}=\frac{5}{6}+\frac{6}{5}=\frac{61}{30}\ne\frac{13}{6}=\text{RHS}$
$\displaystyle \therefore x=\frac{5}{6}\text{ is not a solution.}$
$\displaystyle \text{For }x=\frac{4}{3},$
$\displaystyle \text{LHS}=\frac{4}{3}+\frac{3}{4}=\frac{25}{12}\ne\frac{13}{6}=\text{RHS}$
$\displaystyle \therefore x=\frac{4}{3}\text{ is not a solution.}$

$\displaystyle \text{(iii) For }x=2,$
$\displaystyle \text{LHS}=2(2)^2-2+9=15$
$\displaystyle \text{RHS}=(2)^2+4(2)+3=15$
$\displaystyle \therefore x=2\text{ is a solution.}$
$\displaystyle \text{For }x=3,$
$\displaystyle \text{LHS}=2(3)^2-3+9=24$
$\displaystyle \text{RHS}=(3)^2+4(3)+3=24$
$\displaystyle \therefore x=3\text{ is also a solution.}$

$\displaystyle \text{(iv) For }x=-\sqrt{2},$
$\displaystyle \text{LHS}=(-\sqrt{2})^2-\sqrt{2}(-\sqrt{2})-4=2+2-4=0=\text{RHS}$
$\displaystyle \therefore x=-\sqrt{2}\text{ is a solution.}$
$\displaystyle \text{For }x=-2\sqrt{2},$
$\displaystyle \text{LHS}=(-2\sqrt{2})^2-\sqrt{2}(-2\sqrt{2})-4=8+4-4=8\ne0$
$\displaystyle \therefore x=-2\sqrt{2}\text{ is not a solution.}$

$\displaystyle \text{(v) For }x=\frac{a}{b},$
$\displaystyle \text{LHS}=a^2\left(\frac{a}{b}\right)^2-3ab\left(\frac{a}{b}\right)+2b^2$
$\displaystyle =\frac{a^4}{b^2}-3a^2+2b^2$
$\displaystyle \therefore x=\frac{a}{b}\text{ is not necessarily a solution.}$
$\displaystyle \text{For }x=\frac{b}{a},$
$\displaystyle \text{LHS}=a^2\left(\frac{b}{a}\right)^2-3ab\left(\frac{b}{a}\right)+2b^2$
$\displaystyle =b^2-3b^2+2b^2=0=\text{RHS}$
$\displaystyle \therefore x=\frac{b}{a}\text{ is a solution.}$

$\displaystyle \text{(vi) For }x=3,$
$\displaystyle \text{LHS}=\sqrt{9-12+3}+\sqrt{9-9}=0$
$\displaystyle \text{RHS}=\sqrt{4(9)-14(3)+16}=\sqrt{10}$
$\displaystyle \text{Since LHS}\ne\text{RHS, }x=3\text{ is not a solution.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In each of the following, find the value of }k\text{ for which the given}$
$\displaystyle \text{value is a solution of the given equation:}$
$\displaystyle \text{(i) }7x^2+kx-3=0,\quad x=\frac{2}{3}$
$\displaystyle \text{(ii) }x^2-x(a+b)+k=0,\quad x=a$
$\displaystyle \text{(iii) }kx^2+\sqrt{2}x-4=0,\quad x=\sqrt{2}$
$\displaystyle \text{(iv) }x^2+3ax+k=0,\quad x=-a$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }x=\frac{2}{3}\text{ is a solution, it satisfies the given equation.}$
$\displaystyle 7\left(\frac{2}{3}\right)^2+k\left(\frac{2}{3}\right)-3=0$
$\displaystyle \Rightarrow \frac{28}{9}+\frac{2k}{3}-3=0$
$\displaystyle \Rightarrow \frac{1}{9}+\frac{2k}{3}=0$
$\displaystyle \Rightarrow 1+6k=0$
$\displaystyle \therefore k=-\frac{1}{6}$

$\displaystyle \text{(ii) Since }x=a\text{ is a solution, it satisfies the given equation.}$
$\displaystyle a^2-a(a+b)+k=0$
$\displaystyle \Rightarrow a^2-a^2-ab+k=0$
$\displaystyle \therefore k=ab$

$\displaystyle \text{(iii) Since }x=\sqrt{2}\text{ is a solution, it satisfies the given equation.}$
$\displaystyle k(\sqrt{2})^2+\sqrt{2}(\sqrt{2})-4=0$
$\displaystyle \Rightarrow 2k+2-4=0$
$\displaystyle \Rightarrow 2k=2$
$\displaystyle \therefore k=1$

$\displaystyle \text{(iv) Since }x=-a\text{ is a solution, it satisfies the given equation.}$
$\displaystyle (-a)^2+3a(-a)+k=0$
$\displaystyle \Rightarrow a^2-3a^2+k=0$
$\displaystyle \therefore k=2a^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }x=\frac{2}{3}\text{ and }x=-3\text{ are the roots of the equation}$
$\displaystyle ax^2+7x+b=0,\text{ find the values of }a\text{ and }b.\hfill\text{[CBSE 2016, 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x=\frac{2}{3}\text{ is a root, it satisfies the given equation.}$
$\displaystyle a\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+b=0$
$\displaystyle \Rightarrow \frac{4a}{9}+\frac{14}{3}+b=0$
$\displaystyle \Rightarrow 4a+9b=-42\qquad\ldots\text{(i)}$
$\displaystyle \text{Since }x=-3\text{ is a root, it satisfies the given equation.}$
$\displaystyle a(-3)^2+7(-3)+b=0$
$\displaystyle \Rightarrow 9a+b=21\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (ii), }b=21-9a.$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle 4a+9(21-9a)=-42$
$\displaystyle \Rightarrow 4a+189-81a=-42$
$\displaystyle \Rightarrow -77a=-231$
$\displaystyle \therefore a=3$
$\displaystyle \text{Substituting }a=3\text{ in (ii),}$
$\displaystyle 9(3)+b=21$
$\displaystyle \therefore b=-6$
$\displaystyle \therefore a=3,\quad b=-6$
$\displaystyle \\$