Question

If $2{\sin }^2\frac{\pi }{8}$ is a root of the equation $x^2+ax+b=0$, where $a$ and $b$ are rational numbers, then $a-b$ is equal to

• A. $-\frac{5}{2}$
• B. $-\frac{3}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{5}{2}$

$2{\sin }^2\left(\frac{x}{8}\right)=1-\cos \left(\frac{2x}{8}\right)$ $(\because 2{\sin }^2\theta =1-\cos 2\theta )$

$=1-\cos \left(\frac{\pi }{4}\right)$

$=1-\frac{1}{\sqrt{2}}$

Since one root is $1-\frac{1}{\sqrt{2}}$, therefore other root will be $1+\frac{1}{\sqrt{2}}$

$x^2+ax+b=0$

sum of roots$=-a$

$\therefore 1-\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}=-a$

$\therefore a=-2$

Product of roots $=\frac{c}{a}=b$

$(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})=b$

$1-\frac{1}{2}=b$

$\therefore b=\frac{1}{2}$

$\therefore a-b=-2-\frac{1}{2}=\frac{-5}{2}$.