Let α,β are roots of x2−bx=b(b>0) and |α|,|β| are roots of x2+px+q=0. The minimum value of p2−8qis equal to
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a
-6
b
-4
c
0
d
4
answer is B.
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Detailed Solution
α and β are roots of x2−bx=b∴ α+β=b,αβ=−b|α| and |β| are roots of x2+px+q=0∴ |α|+|β|=−p,|α||β|=q∴ p2−8q=|α|2+|β|2−6|α||β| =(α+β)2−2αβ−6|α||β| =b2+2b+6(−b)=b2−4b=(b−2)2−4 So, minimum value of p2−8q is −4 .