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

Question

Dr. Abubacker · Accepted Answer

(a)
$\because\ 2\sin^{2}\frac{\pi}{8} = 1 - \cos{\frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}}$ (irrational root)
So, other root is $\frac{\sqrt{2} + 1}{\sqrt{2}}.$
Sum of roots $= - a = 1 - \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} = 2 \Rightarrow a = - 2$
Product of roots $= 1 - \frac{1}{2} = \frac{1}{2} = b$
So, $a - b = - 2 - \frac{1}{2} = - \frac{5}{2}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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