If both roots of the equation x^2 - 4x + log_ | vedclass

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Question

(B) For the quadratic equation $x^2 - 4x + \log_{1/2} a = 0$ to have non-real roots,the discriminant $D$ must be less than $0$.$D = b^2 - 4ac < 0$Here

Vedclass · Accepted Answer

$1/16$

Vedclass · Answer

(B) For the quadratic equation $x^2 - 4x + \log_{1/2} a = 0$ to have non-real roots,the discriminant $D$ must be less than $0$.$D = b^2 - 4ac Here,$a = 1$,$b = -4$,and $c = \log_{1/2} a$.$(-4)^2 - 4(1)(\log_{1/2} a) $16 - 4 \log_{1/2} a $16 $4 Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality reverses when we remove the logarithm:$a $a Thus,the maximum value of $a$ is $1/16$ (as $a$ must be strictly less than $1/16$ for the roots to be non-real,the supremum is $1/16$).

If both roots of the equation $x^2 - 4x + \log_{1/2} a = 0$ are not real,what is the maximum value of $a$?

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