Solve: (√36)/(√2·√3) + (√25)/5 | Adding Fractions with Square Roots

Question

Solve the following exercise:

$\frac{\sqrt{36}}{\sqrt{2}\cdot\sqrt{3}}+\frac{\sqrt{25}}{5}=$

00:00 Solve the following problem

00:03 Breakdown 36 into factors of 6 and 6

00:09 When multiplying the square root of a number (A) by the square root of another number (B)

00:12 The result equals the square root of their product (A times B)

00:16 Apply this formula to our exercise

00:31 Breakdown 25 into factors of 5 squared

00:38 Simplify wherever possible

00:43 The square root of any number (A) squared cancels out the square

00:47 Apply this formula to our exercise and cancel out the square

00:53 Any number divided by itself always equals 1

00:56 That's the solution

To solve this problem, let's simplify each term in the expression step-by-step:

Simplify the first term $\frac{\sqrt{36}}{\sqrt{2}\cdot\sqrt{3}}$ :
- $\sqrt{36} = 6$ , as 36 is a perfect square.
- Apply the property of square roots: $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$ .
- Rewrite the expression: $\frac{6}{\sqrt{6}} = \frac{6}{\sqrt{6}}$ .
- Using the square root quotient property: $\frac{6}{\sqrt{6}} = \sqrt{\frac{6^2}{6}} = \sqrt{6}$ .
Simplify the second term $\frac{\sqrt{25}}{5}$ :
- $\sqrt{25} = 5$ , as 25 is a perfect square.
- The expression becomes $\frac{5}{5} = 1$ .
Combine the simplified terms: $\sqrt{6} + 1$

Therefore, the solution to the problem is $\sqrt{6} + 1$ .

$\sqrt{6}+1$