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Transformation of Quadratic Equations - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 12 Questions around this concept.

Solve by difficulty

$
\begin{aligned}
&\text { Find the equation whose roots are reciprocal of the roots of the equation }\\
&a x^2+b x+2=0
\end{aligned}
$

The equation of the smallest degree with real coefficients having $1+i$ as one of the roots is

The number of solutions of the equation $\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$ is:

Concepts Covered - 1

Transformation of Quadratic Equations

Equation from roots

A quadratic equation with $\alpha$ and $\beta$ as its roots is

$$
\begin{aligned}
& (x-\alpha)(x-\beta)=0 \\
& x^2-(\alpha+\beta) x+\alpha \beta=0
\end{aligned}
$$
 

So, the equation with given roots can be written as

x- (sum of roots) x + (Product of roots) = 0

 

Transformation of roots

Let ? and ? be the roots of the quadratic equation $a x^2+b x+c=0$ , then

 

i) the equation with root ? + k and ? + k will be 

$a(x-k)^2+b(x-k)+c=0$,    (replace x by x-k)

 

ii) the equation with root ? - k and ? - k will be 

$a(x+k)^2+b(x+k)+c=0$,  (replace x by x+k)

 

iii) the equation with root ?k and ?k will be 

$a x^2+k b x+k^2 c=0\left(\right.$ replace x by $\left.\frac{x}{k}\right)$

 

iv) the equation with root $\frac{\alpha}{k}$ and $\frac{\beta}{k}$ will be $a x^2 k^2+b k x+c=0$ (replace x by kx )
v) the equation with root -? and -? will be $a x^2-b x+c=0 \quad($ replace x by -x$)$

 

vi) the equation with root $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ will be $c x^2+b x+a=0\left(\right.$ replace x by $\left.\frac{1}{x}\right)$
vii) the equation with root $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$ will be $c x^2-b x+a=0\left(\right.$ replace x by $-\frac{1}{x}$ )
viii) the equation with root $\frac{k}{\alpha}$ and $\frac{k}{\beta}$ will be $c x^2+k b x+k^2 a=0$ (replace x by $\frac{k}{x}$ )

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Transformation of Quadratic Equations

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