Question

Evaluate:

$\sqrt{\sqrt[3]{3125}+\sqrt{24}}$.

• A. $\sqrt{5}-1$
• B. $\sqrt{3}+\sqrt{2}$
• C. $\sqrt{5}+1$
• D. $\sqrt{5}+\sqrt{2}$

Answer 1

Answer: (B)

$\sqrt{3}+\sqrt{2}$

Given, $\sqrt{\sqrt[3]{3125}+\sqrt{24}}$.

$\Rightarrow \sqrt{\sqrt[3]{3125}+\sqrt{24}}=\sqrt{5+\sqrt{24}}$

Let square root of $5+\sqrt{24}$ be $\sqrt{a}+\sqrt{b}$.

Then we can write,

$\Rightarrow 5+\sqrt{24}={(\sqrt{a}+\sqrt{b})}^2=a+b+2\sqrt{ab}$.

Comparing both sides,

$\Rightarrow a+b=\mathrm{5......}\left(1\right)$

$\Rightarrow \sqrt{24}=2\sqrt{ab}$

Squaring both sides,

$\Rightarrow 24=4ab$

$\Rightarrow 6=ab$

$\Rightarrow 6=a(5-a)$

$\Rightarrow 6=5a-a^2$

$\Rightarrow a^2-5a+6=0$

$\Rightarrow a^2-3a-2a+6=0$

$\Rightarrow a(a-3)-2(a-3)=0$

$\Rightarrow (a-2)(a-3)=0$

$\Rightarrow a=3,2$

$\Rightarrow$ for $a=3,b=5-a=5-3=2$

$\Rightarrow \therefore \sqrt{a}+\sqrt{b}=\sqrt{3}+\sqrt{2}$

for $a=2,b=3$

$\therefore \sqrt{a}+\sqrt{b}=\sqrt{2}+\sqrt{3}$

$\therefore \sqrt{5+\sqrt{24}}=\sqrt{2}+\sqrt{3}$.