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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $a$?

Similar Questions

If $\frac{1+\sqrt{3}i}{2}$ is a root of the equation $x^4-x^2+x-1=0$,then its real roots are:

If $\alpha$ and $\beta$ are the least positive integers such that for all $n \in N$, $n^3+\alpha n$ is divisible by $3$ and $n^3-\beta n$ is divisible by $6$, then $\alpha+\beta=$

The equation $x^2-4x+[x]+3=x[x]$,where $[x]$ denotes the greatest integer function,has:

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module