Question

Find the square root of the following surds
12+2root 35
8-2root 15
31-4root 21
30+12root 6

Answer 1

Given:

$\bf 1. \: 12+2\sqrt{35}$

$\bf 2. \: 8-2\sqrt{15}$

$\bf 3. \: 31-4\sqrt{21}$

$\bf 4. \: 30+12\sqrt{6}$

What To Find:

We have to find -

• The square root of the given surds.

How To Find:

To find, we have to -

• First, write in the form of $\sf \sqrt{a}+\sqrt{b}$.
• Next. square the sides by using identities.
• Then, find the square roots of the expression.

Answer 1:

$\sf 12+2\sqrt{35}$

Take it in the form of $\sf \sqrt{a}+\sqrt{b}$ and write it as,

$\sf \to \sqrt{a}+\sqrt{b} = \sqrt{12+2\sqrt{35}}$

Square both sides,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = \bigg(\sqrt{12+2\sqrt{35}}\bigg)^2$

Square the LHS by using the identity,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$

Square the RHS by cancelling the surds and powers,

$\sf \to\bigg(\sqrt{12+2\sqrt{35}}\bigg)^2 = 12+2\sqrt{35}$

After squaring both sides,

$\sf \to a + b + 2\sqrt{ab} = 12 + 2\sqrt{35}$

Let's take -

• a + b = 12
• 2√ab = 2√35

Cancel 2 from both sides,

• √ab = √35

On squaring,

• (√ab)² = (√35)²
• ab - 35

By manual guess,

• a + b = 12 = 7 + 5
• ab = 35 = 7 × 5

∴ Thus, the answer is:-  $\bf \sqrt{7}+\sqrt{5}$.

Answer 2:

$\sf 8-2\sqrt{15}$

Take it in the form of $\sf \sqrt{a}+\sqrt{b}$ and write it as,

$\sf \to \sqrt{a}+\sqrt{b} = \sqrt{8-2\sqrt{15}}$

Square both sides,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = \bigg(\sqrt{8-2\sqrt{15}}\bigg)^2$

Square the LHS by using the identity,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$

Square the RHS by cancelling the surds and powers,

$\sf \to\bigg(\sqrt{8-2\sqrt{15}}\bigg)^2 = 8-2\sqrt{15}$

After squaring both sides,

$\sf \to a + b + 2\sqrt{ab} = 8-2\sqrt{15}$

Let's take -

• a + b = 8
• 2√ab = 2√15

Cancel 2 from both sides,

• √ab = √15

On squaring,

• (√ab)² = (√15)
• ab = 15

By manual guess,

• a + b = 8 = 5 + 3
• ab = 15 = 5 × 3

∴ Thus, the answer is:- $\bf \sqrt{5} + \sqrt{3}$

Answer 3:

$\sf 31-4\sqrt{21}$

Take it in the form of $\sf \sqrt{a}+\sqrt{b}$ and write it as,

$\sf \to \sqrt{a}+\sqrt{b} = \sqrt{31-4\sqrt{21}}$

Square both sides,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = \bigg(\sqrt{31-4\sqrt{21}}\bigg)^2$

Square the LHS by using the identity,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$

Square the RHS by cancelling the surds and powers,

$\sf \to\bigg(\sqrt{31-4\sqrt{21}}\bigg)^2 = 31-4\sqrt{21}$

After squaring both sides,

$\sf \to a + b + 2\sqrt{ab} = 31-4\sqrt{21}$

Let's take -

• a + b = 31
• 2√ab = 4√21

Cancel 2 from both sides,

• √ab = 2√21

Squaring both sides,

• (ab)² = (2√21)²
• ab = 84

By manual guess,

• a + b = 31 = 28 + 3
• ab = 84 = 28 × 3

∴ Thus, the answer is:- $\bf \sqrt{28}+\sqrt{3}$ or  $\bf 2\sqrt{7} + \sqrt{3}$

Answer 4:

$\sf 30+12\sqrt{6}$

Take it in the form of $\sf \sqrt{a}+\sqrt{b}$ and write it as,

$\sf \to \sqrt{a}+\sqrt{b} = \sqrt{30+12\sqrt{6}}$

Square both sides,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = \bigg(\sqrt{30+2\sqrt{6}}\bigg)^2$

Square the LHS by using the identity,

$\sf \to (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$

Square the RHS by cancelling the surds and powers,

$\sf \to\bigg(\sqrt{30+12\sqrt{6}}\bigg)^2 = 30+12\sqrt{6}$

After squaring both sides,

$\sf \to a + b + 2\sqrt{ab} = 30+12\sqrt{6}$

Let's take -

• a + b = 30
• 2√ab = 12√6

Cancel 2 from both sides,

• ab = 6√6

Squaring both sides,

• (√ab)² = (6√6)²
• ab = 216

By manual guess,

• a + b = 30 = 12 + 18
• ab = 216 = 28 × 3

∴ Thus, the answer is:- $\bf \sqrt{18} + \sqrt{12}$ or  $\bf 3\sqrt{2} + 2\sqrt{3}$