Question

If $co{s}^2\frac{\pi }{8}$ is a root of equation $x^2$ + ax + b = 0 where a, b $ϵ$ Q then a + b = ___

• A. $\frac{8}{7}$
• B. $\frac{-7}{8}$
• C. 2
• D. $\frac{-1}{2}$

Answer 1

The correct option is B

$\frac{-7}{8}$

$co{s}^2\frac{\pi }{8}$ is a solution of $x^2$ + ax + b = 0
$\therefore co{s}^4\frac{\pi }{8}+aco{s}^2\frac{\pi }{8}+b=0...\left(1\right)cos\frac{\pi }{4}=2co{s}^2\frac{\pi }{8}-1\Rightarrow co{s}^2\frac{\pi }{8}=(\frac{1}{\sqrt{2}}+1)\frac{1}{2}=co{s}^4\frac{\pi }{8}=(\frac{3}{2}+\sqrt{2})\frac{1}{4}\therefore (\frac{3}{2}+\sqrt{2})\frac{1}{4}+a(\frac{1}{\sqrt{2}}+1)\frac{1}{2}+b=0\text{(substituting in (1) )}\frac{3}{8}+\frac{\sqrt{2}}{4}+\frac{a}{2\sqrt{2}}+\frac{a}{2}+b=0$
$\therefore (\frac{3}{8}+\frac{a}{2}+b)+\sqrt{2}(\frac{1}{4}+\frac{a}{4})=0Sincea,bϵQ\Rightarrow \frac{3}{8}+\frac{a}{2}+b=0\&\frac{a}{4}+\frac{1}{4}=0\frac{3}{8}-\frac{1}{2}+b=0a=-1\therefore b=\frac{1}{8}a+b=-1+\frac{1}{8}=-\frac{7}{8}$