If 2sin2⁡π8 is a root of the equation x2 + ax +b= 0, where a and b are rational numbers, then a - b is equal to

∵2sin2⁡π8=1−cos⁡π4=1−12=2−12 (irrationalroot) So, other root is 2+12sum of roots =−a=1−12+1+12=2⇒a=−2Productof roots =1−12=12=b a−b=−2−12=−52

If 2sin2⁡π8 is a root of the equation x2 + ax +b= 0, where a and b are rational numbers, then a - b is equal to

∵2sin2⁡π8=1−cos⁡π4=1−12=2−12 (irrationalroot) So, other root is 2+12sum of roots =−a=1−12+1+12=2⇒a=−2Productof roots =1−12=12=b a−b=−2−12=−52

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If $2 \sin^{2} \frac{π}{8}$ is a root of the equation x² + ax +b= 0, where a and b are rational numbers, then a - b is equal to

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$- \frac{3}{2}$

$- \frac{5}{2}$

$- \frac{1}{2}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$∵ 2 \sin^{2} \frac{π}{8} = 1 - \cos \frac{π}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} (irrationalroot)$
So, other root is $\frac{\sqrt{2} + 1}{\sqrt{2}}$
sum of roots $= - a = 1 - \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} = 2 \Rightarrow a = - 2$
Productof roots $= 1 - \frac{1}{2} = \frac{1}{2} = b$
$a - b = - 2 - \frac{1}{2} = - \frac{5}{2}$