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Question: If a, b be the roots of the equation *x*<sup>2</sup> – p*x* + q = 0 and a\> 0, b\> 0, then the value...

If a, b be the roots of the equation x2 – px + q = 0 and a> 0, b> 0, then the value of a1/4 + b1/4 is

(p+6q+4q1/4p+2q)k\left( p + 6\sqrt{q} + 4q^{1/4}\sqrt{p + 2\sqrt{q}} \right)^{k}, where k is equal to –

A

1

B

12\frac{1}{2}

C

13\frac{1}{3}

D

14\frac{1}{4}

Answer

14\frac{1}{4}

Explanation

Solution

Since a, b are the roots of the equation

x2 – px + q = 0

\ a + b = p and ab = q

Now, (a 1/4 + b1/4)4 = [(a 1/4 + b1/4)2]2

= [a 1/2 + b1/2 + 2(a b)1/4]2

= [α+β+2αβ+2(αβ)1/4]2\left\lbrack \sqrt{\alpha + \beta + 2\sqrt{\alpha\beta}} + 2(\alpha\beta)^{1/4} \right\rbrack^{2}

= [p+2q+2(q)1/4]2\left\lbrack \sqrt{p + 2\sqrt{q}} + 2(q)^{1/4} \right\rbrack^{2}

= p + 6q\sqrt{q}+ 4q1/4p+2q\sqrt{p + 2\sqrt{q}}

\ a1/4 + b1/4 = [p+6q+4q1/4p+2q]1/4\left\lbrack p + 6\sqrt{q} + 4q^{1/4}\sqrt{p + 2\sqrt{q}} \right\rbrack^{1/4}