Solve (sqrt{5}-sqrt{2})^2+2sqrt{20}divsqrt{2}

$$( \sqrt { 5 } - \sqrt { 2 } ) ^ { 2 } + 2 \sqrt { 20 } \div \sqrt { 2 }$$

Evaluate

$$7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{2}\right)^{2}.

$$\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\frac{2\sqrt{20}}{\sqrt{2}}$$

The square of \sqrt{5} is 5.

$$5-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\frac{2\sqrt{20}}{\sqrt{2}}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$5-2\sqrt{10}+\left(\sqrt{2}\right)^{2}+\frac{2\sqrt{20}}{\sqrt{2}}$$

The square of \sqrt{2} is 2.

$$5-2\sqrt{10}+2+\frac{2\sqrt{20}}{\sqrt{2}}$$

Add 5 and 2 to get 7.

$$7-2\sqrt{10}+\frac{2\sqrt{20}}{\sqrt{2}}$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$7-2\sqrt{10}+\frac{2\times 2\sqrt{5}}{\sqrt{2}}$$

Multiply 2 and 2 to get 4.

$$7-2\sqrt{10}+\frac{4\sqrt{5}}{\sqrt{2}}$$

Rationalize the denominator of \frac{4\sqrt{5}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$7-2\sqrt{10}+\frac{4\sqrt{5}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$7-2\sqrt{10}+\frac{4\sqrt{5}\sqrt{2}}{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$7-2\sqrt{10}+\frac{4\sqrt{10}}{2}$$

Divide 4\sqrt{10} by 2 to get 2\sqrt{10}.

$$7-2\sqrt{10}+2\sqrt{10}$$

Combine -2\sqrt{10} and 2\sqrt{10} to get 0.

$$7$$

Factor

$$7$$