Solve [10-(sqrt{7}-sqrt{3})^2]

$$[ 10 - ( \sqrt { 7 } - \sqrt { 3 } ) ^ { 2 } ]$$

Evaluate

$$2\sqrt{21}\approx 9.16515139$$

Solution Steps

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

$$10-\left(\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$10-\left(7-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

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

$$10-\left(7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$10-\left(7-2\sqrt{21}+3\right)$$

Add 7 and 3 to get 10.

$$10-\left(10-2\sqrt{21}\right)$$

To find the opposite of 10-2\sqrt{21}, find the opposite of each term.

$$10-10+2\sqrt{21}$$

Subtract 10 from 10 to get 0.

$$2\sqrt{21}$$

Expand

$$2 \sqrt{21} = 9.16515139$$

Solution Steps

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

$$10-\left(\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$10-\left(7-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

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

$$10-\left(7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$10-\left(7-2\sqrt{21}+3\right)$$

Add 7 and 3 to get 10.

$$10-\left(10-2\sqrt{21}\right)$$

To find the opposite of 10-2\sqrt{21}, find the opposite of each term.

$$10-10+2\sqrt{21}$$

Subtract 10 from 10 to get 0.

$$2\sqrt{21}$$