Find the value of a 5th-root expression.

Question

Simplify and find the value of the expression:

$$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}.$$

I tried to rationalise it.

It was of no use..

Answer 1

Let $$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}=x$$

Raise to the power of 5 on both sides, $$(\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}})^5=x^5$$

Use the identity: $$(a+b)^5=a^5+b^5+5ab[(a+b)^3-ab(a+b)]$$ We know that $(a+b)=x$.

Simlifying $a×b$ that is $\sqrt[5]{\frac{123+\sqrt{15125}}{2}}×\sqrt[5]{\frac{123-\sqrt{15125}}{2}},$ we get $1$.

By substituting, the expression simplifies to the polynomial $$x^5=123+5[x^3-x]$$ $$x^5-5x^3+5x-123=0$$ Solving the degree 5 polynomial, we get $$x=3$$..$Ans..$

Answer 2

$$\frac{123+\sqrt{15125}}2=\frac{123+55\sqrt5}2=\left(\frac{1+\sqrt5}2\right)^{10}=\left(\frac{3+\sqrt5}2\right)^5.$$ Likewise $$\frac{123-\sqrt{15125}}2=\left(\frac{3-\sqrt5}2\right)^5.$$ So $$\sqrt[5]{\frac{123+\sqrt{15125}}2}+\sqrt[5]{\frac{123-\sqrt{15125}}2} =\frac{3+\sqrt5}2+\frac{3-\sqrt5}2=3.$$