Solve (sqrt{7}+3)^2-(sqrt{14}-sqrt{2})^2

$$( \sqrt { 7 } + 3 ) ^ { 2 } - ( \sqrt { 14 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$10\sqrt{7}\approx 26.457513111$$

Solution Steps

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

$$\left(\sqrt{7}\right)^{2}+6\sqrt{7}+9-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$7+6\sqrt{7}+9-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

Add 7 and 9 to get 16.

$$16+6\sqrt{7}-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{14}-\sqrt{2}\right)^{2}.

$$16+6\sqrt{7}-\left(\left(\sqrt{14}\right)^{2}-2\sqrt{14}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{14} is 14.

$$16+6\sqrt{7}-\left(14-2\sqrt{14}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

Factor 14=2\times 7. Rewrite the square root of the product \sqrt{2\times 7} as the product of square roots \sqrt{2}\sqrt{7}.

$$16+6\sqrt{7}-\left(14-2\sqrt{2}\sqrt{7}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$16+6\sqrt{7}-\left(14-2\times 2\sqrt{7}+\left(\sqrt{2}\right)^{2}\right)$$

Multiply -2 and 2 to get -4.

$$16+6\sqrt{7}-\left(14-4\sqrt{7}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$16+6\sqrt{7}-\left(14-4\sqrt{7}+2\right)$$

Add 14 and 2 to get 16.

$$16+6\sqrt{7}-\left(16-4\sqrt{7}\right)$$

To find the opposite of 16-4\sqrt{7}, find the opposite of each term.

$$16+6\sqrt{7}-16+4\sqrt{7}$$

Subtract 16 from 16 to get 0.

$$6\sqrt{7}+4\sqrt{7}$$

Combine 6\sqrt{7} and 4\sqrt{7} to get 10\sqrt{7}.

$$10\sqrt{7}$$

Expand

$$10 \sqrt{7} = 26.457513111$$

Solution Steps

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

$$\left(\sqrt{7}\right)^{2}+6\sqrt{7}+9-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

The square of \sqrt{7} is 7.

$$7+6\sqrt{7}+9-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

Add 7 and 9 to get 16.

$$16+6\sqrt{7}-\left(\sqrt{14}-\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{14}-\sqrt{2}\right)^{2}.

$$16+6\sqrt{7}-\left(\left(\sqrt{14}\right)^{2}-2\sqrt{14}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{14} is 14.

$$16+6\sqrt{7}-\left(14-2\sqrt{14}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

Factor 14=2\times 7. Rewrite the square root of the product \sqrt{2\times 7} as the product of square roots \sqrt{2}\sqrt{7}.

$$16+6\sqrt{7}-\left(14-2\sqrt{2}\sqrt{7}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$16+6\sqrt{7}-\left(14-2\times 2\sqrt{7}+\left(\sqrt{2}\right)^{2}\right)$$

Multiply -2 and 2 to get -4.

$$16+6\sqrt{7}-\left(14-4\sqrt{7}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$16+6\sqrt{7}-\left(14-4\sqrt{7}+2\right)$$

Add 14 and 2 to get 16.

$$16+6\sqrt{7}-\left(16-4\sqrt{7}\right)$$

To find the opposite of 16-4\sqrt{7}, find the opposite of each term.

$$16+6\sqrt{7}-16+4\sqrt{7}$$

Subtract 16 from 16 to get 0.

$$6\sqrt{7}+4\sqrt{7}$$

Combine 6\sqrt{7} and 4\sqrt{7} to get 10\sqrt{7}.

$$10\sqrt{7}$$