Question

If one root of the equation $2x^2-ax+64=0$ is twice the other, then find the value of a.

• A. $a=\pm 24$
• B. $a=\pm 4$
• C. $a=\pm 2$
• D. None of these

Answer 1

Answer: (A)

$a=\pm 24$

$\Rightarrow$ Let $\alpha$ and $\beta$ are roots of the equation, $2x^2-ax+64=0$

$\Rightarrow$ Here, $a=2,b=-a$ and $c=64$

According to the question,

$\Rightarrow$ $\beta =2\alpha$

$\Rightarrow$ $\alpha .\beta =\frac{c}{a}$

$\Rightarrow$ $\alpha .2\alpha =\frac{64}{2}$

$\Rightarrow$ $2{\alpha }^2=32$

$\Rightarrow$ ${\alpha }^2=16$

$\therefore$ $\alpha =\pm 4$

Put $x=4$

$\Rightarrow$ $2(4{)}^2-a(4)+64=0$

$\Rightarrow$ $4a=96$

$\therefore$ $a=24$

Put $x=-4$

$\Rightarrow$ $2(-4{)}^2-a(-4)+64=0$

$\Rightarrow$ $4a=-96$

$\therefore$ $a=-24$

$\therefore$ $a=\pm 24$