Question

The equation whose roots are twice the roots of the equation $x^2-3x+3=0$ is

• A. $x^2-3x+6=0$
• B. $x^2-4x+8=0$
• C. $x^2-6x+12=0$
• D. $x^2-8x+6=0$

Answer 1

The correct option is C

$x^2-6x+12=0$

Find the equation:

Let $\alpha$ and $\beta$ are the roots of the given equation $x^2-3x+3=0$. Then,

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

$\begin{array}{rcl}& \Rightarrow & \alpha +\beta =-\frac{\left(-3\right)}{1}\\ & =& 3...\left(1\right)\end{array}$

And

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

$\begin{array}{rcl}& \Rightarrow & \alpha ·\beta =\frac{3}{1}\\ & =& 3...\left(2\right)\end{array}$

The roots of the required equation are $2\alpha$ and $2\beta$. [Given]

Now,

Sum of roots$=2\alpha +2\beta$

$\begin{array}{rcl}& \Rightarrow & 2\alpha +2\beta =2\left(\alpha +\beta \right)\\ & =& 2\times 3\left[\mathrm{from}\left(1\right)\right]\\ & =& 6\end{array}$

Product of roots$=2\alpha ·2\beta$

$\begin{array}{rcl}& \Rightarrow & 2\alpha ·2\beta =4\alpha ·\beta \\ & =& 4\times 3\left[\mathrm{from}\left(2\right)\right]\\ & =& 12\end{array}$

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

$\Rightarrow x^2-6x+12=0$

Hence, the correct option is C.