Question

If $1,5,7$ are the roots of $a{x}^3+b{x}^2+cx+d=0$, then the roots of $a{x}^3+2b{x}^2+4cx+8d=0$ are:

• A. $2,10,14$
• B. $\frac{1}{2},\frac{5}{2},\frac{7}{2}$
• C. $-2,-10,-14$
• D. $0,2,10$

Answer 1

The correct option is A $2,10,14$

Dividing the whole equation by $8$, we get,
$a{\left(\frac{x}{2}\right)}^3+b{\left(\frac{x}{2}\right)}^2+c\left(\frac{x}{2}\right)+d=0$

We know that , $\frac{x}{2}=1,5,7$; i.e. $x=2,10,14$ are solutions of these by comparing this with the above equation.