Solve (8sqrt{3}+sqrt{7})^2

$$( 8 \sqrt { 3 } + \sqrt { 7 } ) ^ { 2 }$$

Evaluate

$$16\sqrt{21}+199\approx 272.321211119$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8\sqrt{3}+\sqrt{7}\right)^{2}.

$$64\left(\sqrt{3}\right)^{2}+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{3} is 3.

$$64\times 3+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

Multiply 64 and 3 to get 192.

$$192+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$192+16\sqrt{21}+\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$192+16\sqrt{21}+7$$

Add 192 and 7 to get 199.

$$199+16\sqrt{21}$$

Expand

$$16 \sqrt{21} + 199 = 272.321211119$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8\sqrt{3}+\sqrt{7}\right)^{2}.

$$64\left(\sqrt{3}\right)^{2}+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{3} is 3.

$$64\times 3+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

Multiply 64 and 3 to get 192.

$$192+16\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$192+16\sqrt{21}+\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$192+16\sqrt{21}+7$$

Add 192 and 7 to get 199.

$$199+16\sqrt{21}$$