The square root of sqrt{50} + sqrt{48} is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.We want to find $\sqrt{\sqrt{50} + \sqrt{48}}$.Note that $\sqrt{50} + \sqrt{48}

Vedclass · Accepted Answer

$2^{1/4}(\sqrt{3} + \sqrt{2})$

Vedclass · Answer

(D) Given expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.We want to find $\sqrt{\sqrt{50} + \sqrt{48}}$.Note that $\sqrt{50} + \sqrt{48} = \sqrt{2}(25 + \sqrt{24}) = \sqrt{2}(5^2 + (\sqrt{24})^2 + 2 	imes 5 	imes \sqrt{24})$ is not quite right. Let us rewrite $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.Consider the square of $(\sqrt{3} + \sqrt{2})$: $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.Then $\sqrt{2}(\sqrt{3} + \sqrt{2})^2 = \sqrt{2}(5 + 2\sqrt{6}) = 5\sqrt{2} + 2\sqrt{12} = 5\sqrt{2} + 4\sqrt{3} = \sqrt{50} + \sqrt{48}$.Therefore,$\sqrt{\sqrt{50} + \sqrt{48}} = \sqrt{\sqrt{2}(\sqrt{3} + \sqrt{2})^2} = 2^{1/4}(\sqrt{3} + \sqrt{2})$.

The square root of $\sqrt{50} + \sqrt{48}$ is

Similar Questions

What is the square root of $\sqrt{50} + \sqrt{48}$?

The imaginary part of $i^{i}$ is

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

If $\tanh^{-1}(x+iy) = \frac{1}{2} \tanh^{-1}\left(\frac{2x}{1+x^2+y^2}\right) + \frac{i}{2} \tan^{-1}\left(\frac{2y}{1-x^2-y^2}\right)$,where $x, y \in \mathbb{R}$,then $\tanh^{-1}(iy) =$

If $\sqrt{a + ib} = x + iy$,then the possible value of $\sqrt{a - ib}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module