Question

If $\alpha ,\beta$ are the roots of the equation $x^2-px+q=0and\alpha$ > $0,\beta$ >$0$ , then the value of ${\alpha }^{\frac{1}{4}}+{\beta }^{\frac{1}{4}}is{(p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}})}^k$ , where k is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$x^2-px+q=0\therefore \alpha +\beta =pand\alpha \beta =q.$

$Now,{({\alpha }^{\frac{1}{4}}+{\beta }^{\frac{1}{4}})}^4={\left[{({\alpha }^{\frac{1}{4}}+{\beta }^{\frac{1}{4}})}^2\right]}^2={[{\alpha }^{\frac{1}{2}}+{\beta }^{\frac{1}{2}}+(\alpha \beta {)}^{\frac{1}{4}}]}^2$

$={[\sqrt{\alpha +\beta +2\sqrt{\alpha \beta }}+2(\alpha \beta {)}^{\frac{1}{4}}]}^2={[\sqrt{p+2\sqrt{q}}+2(q{)}^{\frac{1}{4}}]}^2=p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}}\therefore {\alpha }^{\frac{1}{4}}+{\beta }^{\frac{1}{4}}={[p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}}]}^{\frac{1}{4}}$