Question

If $\alpha ,\beta$ are the roots of the quadratic equation $x^2-14x+45=0,$ then the quadratic equation with roots as $3\alpha ,3\beta$ is

• A. $\frac{x^2}{9}-\frac{14x}{3}+5=0$
• B. $9x^2-42x+405=0$
• C. $x^2-42x+405=0$

Answer 1

The correct option is C $x^2-42x+405=0$

Given: $\alpha ,\beta$ as the roots of the $x^2-14x+45=0$
To find: Quadratic equation with roots as $3\alpha ,3\beta$

Now, we remember if the roots $p,q$ of any quadratic equation $a{x}^2+bx+c=0$ are transformed to $kp,kq\forall k\in R$
Then the transformed equation with roots as $kp,kq$ is given as:
$a(\frac{x}{k}{)}^2+b(\frac{x}{k})+c=0$
Thus using the same method, we get our transformed quadratic equation as:
$(\frac{x}{3}{)}^2-14(\frac{x}{3})+45=0\Rightarrow x^2-14\times 3\times x+45\times 9=0\Rightarrow x^2-42x+405=0$
hence, the transformed equation is $x^2-42x+405=0$