Question

Which one of the following is the equation whose roots are respectively three times the roots of the equation $a{x}^2+bx+c=0$?

• A. $a{x}^2+3bx+c=0$
• B. $a{x}^2+3bx+9c=0$
• C. $a{x}^2-3bx+9c=0$
• D. $a{x}^2+bx+3c=0$

Answer 1

The correct option is B

$a{x}^2+3bx+9c=0$

Explanation for the correct option:

Find the quadratic equation:

Let $\alpha ,\beta$ be the roots of the equation $a{x}^2+bx+c=0$. Then

Sum of root$=-\frac{\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}{x}^2}$

$\Rightarrow \alpha +\beta =-\frac{b}{a}...\left(1\right)$

Product of root$=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$

$\Rightarrow \alpha ·\beta =\frac{c}{a}...\left(2\right)$

By the given condition, roots of the required equation are $3\alpha$ and $3\beta$.

Now,

Sum of root$=3\alpha +3\beta$

$\begin{array}{rcl}& \Rightarrow & 3\alpha +3\beta =3\left(\alpha +\beta \right)\\ & =& -\frac{3b}{a}...\left(3\right)\left[\mathrm{from}\left(1\right)\right]\end{array}$

And

Product of root$=3\alpha ·3\beta$

$\begin{array}{rcl}& \Rightarrow & 3\alpha ·3\beta =9\alpha \beta \\ & =& \frac{9c}{a}...\left(4\right)\left[\mathrm{from}\left(2\right)\right]\end{array}$

The required equation is $x^2-\left(\mathrm{sum}\mathrm{of}\mathrm{root}\right)x+\left(\mathrm{product}\mathrm{of}\mathrm{root}\right)=0$

Substitute equation $\left(3\right)$ and $\left(4\right)$ in the required equation, we get

$$\begin{gathered} \Rightarrow x^2-\left(\frac{-3b}{a}\right)x+\frac{9c}{a}=0 \\ \Rightarrow a{x}^2+3bx+9c=0 \end{gathered}$$

Hence, the correct option is B.