√(3+√5)= Option 1: √2+1 Option 2: √(5/2)+√(1/2) Option 3: √(7/2)-√(1/2) Option 4: √(9/2)-√(3/2)

Answer: √(5/2)+√(1/2). Solution: To solve this easily, we can go through the options and square the given options to see if they equate 3+5 Look at the second option. We know that (a+b)^2=a^2+b^2+2ab Squaring the second option, we'll get: 5/2+1/2+2(√(5/2))(√(1/2)) 6/2+(25)/2=3+5

Algebra

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Medium

$\sqrt{3+\sqrt{5}}=$

$\sqrt{2}+1$

$\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}$

$\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}$

$\sqrt{\frac{9}{2}}-\sqrt{\frac{3}{2}}$

✅ Correct Option: 2

To solve this easily, we can go through the options and square the given options to see if they equate $3+\sqrt5$

Look at the second option.

We know that $(a+b)^2=a^2+b^2+2ab$

Squaring the second option, we'll get:

$\frac{5}{2}+\frac{1}{2}+2(\sqrt{\frac{5}{2}})(\sqrt{\frac{1}{2}})$

$\frac{6}{2}+\frac{2\sqrt5}{2}=3+\sqrt5$

JIPMAT 2021

If $2^{x} = 3^{y} = 6^{-z}$ , then $(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is equal to

JIPMAT 2025

$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

JIPMAT 2024

Simplify : $\dfrac{\sqrt[4]{0.0625}+\sqrt[3]{0.008}+\sqrt{0.09}-1}{\sqrt[3]{62 \cdot 5 \times \sqrt[5]{32}}}$

JIPMAT 2024

Choose the correct answer from the options given below :

JIPMAT 2026

The value of $3+\frac{1}{\sqrt3}+\frac{1}{3+\sqrt3}+\frac{1}{\sqrt3-3}$ is

JIPMAT 2026

Arrange the following in descending order:

A. $\sqrt{401}-\sqrt{399}$

B. $\sqrt{301}-\sqrt{299}$

C. $\sqrt{101}-\sqrt{99}$

D. $\sqrt{201}-\sqrt{199}$

Choose the correct answer from the options given below:

3+5=\sqrt{3+\sqrt{5}}=3+5​​=

Solution

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$\sqrt{3+\sqrt{5}}=$