Question

If one root of the equation $a{x}^2+bx+c=0$ is double of the other, then $2b^2$ is equal to

• A. 9 ca
• B. $c\sqrt{2}a$
• C. $2\sqrt{3}ac$
• D. None of these

Answer 1

The correct option is A 9 ca

Let the roots of the given equation be 1 and 2.
Then the equation $a{x}^2+bx+c=0$ becomes $x^2-3x+2=0$
$\therefore$ a = 1, b = -3, c = 2.
Now, $2b^2=2\times (-3{)}^2=18$
Check options which gives 18 when we put values for a and c as 1 and 2 respectively.
Option (a) is 9ac = $9\times 1\times 2=18$
Hence, option (a) is correct answer.

Alternatively:
Let the roots of the given equation are $\alpha \beta$.
Given $\alpha =2\beta$ . . . (1)
Given equation be $x^2$ + bx + c = 0
Sum of the roots $(\alpha +\beta )=-\frac{b}{a}$
and product of roots $(\alpha \times \beta )=\frac{c}{a}$
From $\alpha +\beta =-\frac{b}{a}$
$2\beta +\beta =-\frac{b}{a}$ [by Equation (1)]
$3\beta =-\frac{b}{a}\Rightarrow \beta =-\frac{b}{3a}$
${\beta }^2=\frac{b^2}{9a^2}$ . . . (2)
Now, $\alpha \times \beta =\frac{c}{a}$
$2\beta \times \beta =\frac{c}{a}\Rightarrow 2{\beta }^2=\frac{c}{a}$
$={\beta }^2=\frac{c}{2a}$ . . . (3)
Fron Eqs. (2) and (3) $\frac{b^2}{9a^2}=\frac{c}{2a}$
$=2b^2=\frac{9a^{2}c}{a}=9ca=2b^2=9ca$