Solve (2sqrt{7}+7sqrt{2})^2-(2sqrt{7}-7sqrt{5})

$$( 2 \sqrt { 7 } + 7 \sqrt { 2 } ) ^ { 2 } - ( 2 \sqrt { 7 } - 7 \sqrt { 5 } )$$

Evaluate

$$7\sqrt{5}+28\sqrt{14}+126-2\sqrt{7}\approx 241.12738005$$

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}.

$$4\left(\sqrt{7}\right)^{2}+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}-7\sqrt{5}\right)$$

The square of \sqrt{7} is 7.

$$4\times 7+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}-7\sqrt{5}\right)$$

Multiply 4 and 7 to get 28.

$$28+28\sqrt{7}\sqrt{2}+49\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}-7\sqrt{5}\right)$$

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

$$28+28\sqrt{14}+49\left(\sqrt{2}\right)^{2}-\left(2\sqrt{7}-7\sqrt{5}\right)$$

The square of \sqrt{2} is 2.

$$28+28\sqrt{14}+49\times 2-\left(2\sqrt{7}-7\sqrt{5}\right)$$

Multiply 49 and 2 to get 98.

$$28+28\sqrt{14}+98-\left(2\sqrt{7}-7\sqrt{5}\right)$$

Add 28 and 98 to get 126.

$$126+28\sqrt{14}-\left(2\sqrt{7}-7\sqrt{5}\right)$$

To find the opposite of 2\sqrt{7}-7\sqrt{5}, find the opposite of each term.

$$126+28\sqrt{14}-2\sqrt{7}+7\sqrt{5}$$