If $\alpha, \beta$ be the positive real roots of the equation $x^2-p x+q=0$, then the value of $\alpha^{1 / 4}+\beta^{1 / 4}$ is $\left(p+6 \sqrt{q}+4 q^{1 / 4} \sqrt{p+2 \sqrt{q}}\right)^k$, where $\mathrm{k}$ is equal to

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

\alpha, \beta

be the positive real roots of the equation

x^2-p x+q=0

, then the value of

\alpha^{1 / 4}+\beta^{1 / 4}

\left(p+6 \sqrt{q}+4 q^{1 / 4} \sqrt{p+2 \sqrt{q}}\right)^k

, where

\mathrm{k}

is equal to

View answer and explanation

Correct option: D

\frac{1}{4}

Option D (1/4) is correct because after simplifying the expression

(\alpha^{1/4} + \beta^{1/4})^4

, we find that k = 1/4.

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