Solve {left(3sqrt{7}+2sqrt{2}right)}^2

$${ \left(3 \sqrt{ 7 } +2 \sqrt{ 2 } \right) }^{ 2 }$$

Evaluate

$$12\sqrt{14}+71\approx 115.899888641$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3\sqrt{7}+2\sqrt{2}\right)^{2}.

$$9\left(\sqrt{7}\right)^{2}+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$9\times 7+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

Multiply 9 and 7 to get 63.

$$63+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$63+12\sqrt{14}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$63+12\sqrt{14}+4\times 2$$

Multiply 4 and 2 to get 8.

$$63+12\sqrt{14}+8$$

Add 63 and 8 to get 71.

$$71+12\sqrt{14}$$

Expand

$$12 \sqrt{14} + 71 = 115.899888641$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3\sqrt{7}+2\sqrt{2}\right)^{2}.

$$9\left(\sqrt{7}\right)^{2}+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$9\times 7+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

Multiply 9 and 7 to get 63.

$$63+12\sqrt{7}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$63+12\sqrt{14}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$63+12\sqrt{14}+4\times 2$$

Multiply 4 and 2 to get 8.

$$63+12\sqrt{14}+8$$

Add 63 and 8 to get 71.

$$71+12\sqrt{14}$$