Solve {left(sqrt{6}+6sqrt{7}right)}^2

$${ \left( \sqrt{ 6 } +6 \sqrt{ 7 } \right) }^{ 2 }$$

Evaluate

$$12\sqrt{42}+258\approx 335.768888381$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+6\sqrt{7}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{7}, multiply the numbers under the square root.

$$6+12\sqrt{42}+36\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$6+12\sqrt{42}+36\times 7$$

Multiply 36 and 7 to get 252.

$$6+12\sqrt{42}+252$$

Add 6 and 252 to get 258.

$$258+12\sqrt{42}$$

Expand

$$12 \sqrt{42} + 258 = 335.768888381$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+6\sqrt{7}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{7}, multiply the numbers under the square root.

$$6+12\sqrt{42}+36\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$6+12\sqrt{42}+36\times 7$$

Multiply 36 and 7 to get 252.

$$6+12\sqrt{42}+252$$

Add 6 and 252 to get 258.

$$258+12\sqrt{42}$$