Question

If $\alpha ,\beta$ are roots of equation $a{x}^2+bx+c=0$ which are real and opposite in sign, then roots of the equation $\alpha {(x-\beta )}^2+\beta {(x-\alpha )}^2=0$ are

• A. real and opposite in signs
• B. imaginary
• C. positive
• D. negative

Answer 1

The correct option is C positive

$a{x}^2+bx+c=0$
Therefore, $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$
and $D=b^2-4ac>0$
Now, $\alpha (x-\beta {)}^2+\beta (x-\alpha {)}^2=0$
$\Rightarrow (\alpha +\beta )x^2-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0$
$\Rightarrow -\frac{b{x}^2}{a}-\frac{4cx}{a}-\frac{bc}{a^2}=0$
$\Rightarrow ab{x}^2+4acx+bc=0$
Since, $\alpha \beta <0$ and $D>0$
Therefore, $D_1>0$
Now, product of the roots of the equation $=\frac{bc}{ab}=\alpha \beta <0$
Therefore, roots are of opposite sign.
Ans: A