If $\alpha,\ \beta$ are the roots of the equation $ax^2+bx+c=0$, then find the roots of the equation $ax^2+bx(x+1)+c(x+1)^2=0$.

Given: $\alpha,\ \beta$ are the roots of the equation $ax^2+bx+c=0$.

To do: To find the roots of the equation $ax^2+bx(x+1)+c(x+1)^2=0$.

Solution:

As given, $\alpha$ and $\beta$ are the roots of equation $ax^2+bx+c=0$.

$\Rightarrow \alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$

Now for the equation is $ax^2+bx(x+1)+c(x+1)^2=0$.

Let $p$ and $q$ are the roots of the equation.

$\Rightarrow ax^2+bx^2+bx+cx^2+2cx+c=0$

$\Rightarrow ( a+b+c)x^2+( b+2c)x+c=0$

$\Rightarrow$ Sum of the roots $p+q=-\frac{b+2c}{a+b+c}$

$\Rightarrow p+q=-\frac{\frac{b}{a}+\frac{2c}{a}}{\frac{a}{a}+\frac{b}{a}+\frac{c}{a}}$

$\Rightarrow p+q=-\frac{-( \alpha+\beta)+2\alpha\beta}{1-( \alpha+\beta)+\alpha\beta}$

$\Rightarrow p+q=\frac{\alpha+\beta-2\alpha\beta}{1-( \alpha+\beta)+\alpha\beta}$

$\Rightarrow p+q=\frac{\alpha-\alpha\beta+\beta-\alpha\beta}{1-\alpha-\beta+\alpha\beta}$

$\Rightarrow p+q=\frac{\alpha( 1-\beta)+\beta( 1-\alpha)}{( 1-\alpha)( 1-\beta)}$

$\Rightarrow p+q=\frac{\alpha( 1-\beta)}{( 1-\alpha)( 1-\beta)}+\frac{\beta( 1-\alpha)}{( 1-\alpha)( 1-\beta)}$

$\Rightarrow p+q=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}\ .........\ ( i)$

And Product of the roots $pq=\frac{c}{a+b+c}$

$\Rightarrow pq=\frac{\frac{c}{a}}{\frac{a}{a}+\frac{b}{a}+\frac{c}{a}}$

$\Rightarrow pq=\frac{\alpha\beta}{1-( \alpha+\beta)+\alpha\beta}$

$\Rightarrow pq=\frac{\alpha\beta}{( 1-\alpha)( 1-\beta)}$

$\Rightarrow pq=\frac{\alpha}{1-\alpha}.\frac{\beta}{1-\beta}\ .......\ ( ii)$

From $( i)$ and $( ii)$

$p=\frac{\alpha}{1-\alpha}$ and $q=\frac{\beta}{1-\beta}$

Thus, the roots of the equation $ax^2+bx(x+1)+c(x+1)^2=0$ are $p=\frac{\alpha}{1-\alpha}$ and $q=\frac{\beta}{1-\beta}$.

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Updated on: 10-Oct-2022

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