Question

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0,a\ne 0$ and ${\alpha }_1,-\beta$ be the roots of $a_1{x}^2+b_{1}x+c_1=0,a_1\ne 0$. Then the quadratic equation whose roots are $\alpha ,{\alpha }_1$ is

• A. $\frac{x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}-x+\frac{1}{(\frac{b}{c}+\frac{b_1}{c_1})}=0$
• B. $\frac{x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}+2x+\frac{1}{(\frac{b}{c}-\frac{b_1}{c_1})}=0$
• C. $\frac{2x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}+x+\frac{1}{(\frac{b}{c}+\frac{b_1}{c_1})}=0$
• D. $\frac{x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}+x+\frac{1}{(\frac{b}{c}+\frac{b_1}{c_1})}=0$

Answer 1

The correct option is D $\frac{x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}+x+\frac{1}{(\frac{b}{c}+\frac{b_1}{c_1})}=0$

$\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$
$\Rightarrow \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
${\alpha }_1,-\beta$ are the roots of $a_1{x}^2+b_{1}x+c_1=0$
$\Rightarrow {\alpha }_1-\beta =-\frac{b_1}{a_1},-{\alpha }_1\beta =\frac{c_1}{a_1}$

Now, $(\alpha +\beta )+({\alpha }_1-\beta )=-\frac{b}{a}-\frac{b_1}{a_1}=\alpha +{\alpha }_1$
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-b/a}{c/a}=-\frac{b}{c}\cdots \left(1\right)$
$\frac{1}{{\alpha }_1}-\frac{1}{\beta }=\frac{\beta -{\alpha }_1}{{\alpha }_1\beta }=\frac{-b_1}{c_1}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$\frac{1}{\alpha }+\frac{1}{{\alpha }_1}=-(\frac{b}{c}+\frac{b_1}{c_1})=\frac{\alpha +{\alpha }_1}{\alpha {\alpha }_1}$
$\therefore \alpha {\alpha }_1=\frac{\frac{b}{a}+\frac{b_1}{a_1}}{\frac{b}{c}+\frac{b_1}{c_1}}$

The equation whose roots are $\alpha ,{\alpha }_1$ is
$x^2-(\alpha +{\alpha }_1)x+\alpha {\alpha }_1=0$
$\Rightarrow x^2+(\frac{b}{a}+\frac{b_1}{a_1})x+\frac{\frac{b}{a}+\frac{b_1}{a_1}}{\frac{b}{c}+\frac{b_1}{c_1}}=0$
$\Rightarrow \frac{x^2}{(\frac{b}{a}+\frac{b_1}{a_1})}+x+\frac{1}{(\frac{b}{c}+\frac{b_1}{c_1})}=0$