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

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

Evaluate

$$28\sqrt{14}+126\approx 230.76640683$$

Solution Steps

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

$$4\left(\sqrt{7}\right)^{2}+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$4\times 7+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

Multiply 4 and 7 to get 28.

$$28+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

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

$$28+28\sqrt{14}+49\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$28+28\sqrt{14}+49\times 2$$

Multiply 49 and 2 to get 98.

$$28+28\sqrt{14}+98$$

Add 28 and 98 to get 126.

$$126+28\sqrt{14}$$

Expand

$$28 \sqrt{14} + 126 = 230.76640683$$

Solution Steps

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

$$4\left(\sqrt{7}\right)^{2}+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$4\times 7+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

Multiply 4 and 7 to get 28.

$$28+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}$$

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

$$28+28\sqrt{14}+49\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$28+28\sqrt{14}+49\times 2$$

Multiply 49 and 2 to get 98.

$$28+28\sqrt{14}+98$$

Add 28 and 98 to get 126.

$$126+28\sqrt{14}$$