If $\alpha\ and\ \beta$ are the roots of the equation $ax^{2} + bx + c = 0,\ (c
eq 0)$, then the equation whose roots are $\frac{1}{a\alpha + b}$ and $\frac{1}{a\beta + b}$ is

Question

If $\alpha\ and\ \beta$ are the roots of the equation $ax^{2} + bx + c = 0,\ (c 
eq 0)$, then the equation whose roots are $\frac{1}{a\alpha + b}$ and $\frac{1}{a\beta + b}$ is

Dr. Abubacker · Accepted Answer

(a)
Replacing $x$ by $\frac{1 - bx}{ax}$ we get the required equation
$a\left( \frac{1 - bx}{ax} \right)^{2} + b\left( \frac{1 - bx}{ax} \right) + c = 0$
$\Rightarrow a\left( 1 + b^{2}x^{2} - 2bx \right) + ax\left( b - b^{2}x \right) + ca^{2}x^{2} = 0$
$\Rightarrow a + ab^{2}x^{2} - 2abx + abx - ab^{2}x^{2} + a^{2}cx^{2} = 0$
$\Rightarrow \ \ \ \ \ \ acx^{2} - bx + 1 = 0$

Q. If $\alpha\ and\ \beta$ are the roots of the equation $ax^{2} + bx + c = 0,\ (c \neq 0)$, then the equation whose roots are $\frac{1}{a\alpha + b}$ and $\frac{1}{a\beta + b}$ is

Explanation

Reviewed by Dr. Abubacker

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