Solve (sqrt{8}-sqrt{40}+sqrt{18})^2

$$( \sqrt { 8 } - \sqrt { 40 } + \sqrt { 18 } ) ^ { 2 }$$

Evaluate

$$90-40\sqrt{5}\approx 0.5572809$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}-\sqrt{40}+\sqrt{18}\right)^{2}$$

Factor 40=2^{2}\times 10. Rewrite the square root of the product \sqrt{2^{2}\times 10} as the product of square roots \sqrt{2^{2}}\sqrt{10}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}-2\sqrt{10}+\sqrt{18}\right)^{2}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

Combine 2\sqrt{2} and 3\sqrt{2} to get 5\sqrt{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5\sqrt{2}-2\sqrt{10}\right)^{2}.

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$25\times 2-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$50-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$50-20\sqrt{2}\sqrt{2}\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

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

$$50-20\times 2\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

Multiply -20 and 2 to get -40.

$$50-40\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$50-40\sqrt{5}+4\times 10$$

Multiply 4 and 10 to get 40.

$$50-40\sqrt{5}+40$$

Add 50 and 40 to get 90.

$$90-40\sqrt{5}$$

Expand

$$90-40\sqrt{5}$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}-\sqrt{40}+\sqrt{18}\right)^{2}$$

Factor 40=2^{2}\times 10. Rewrite the square root of the product \sqrt{2^{2}\times 10} as the product of square roots \sqrt{2^{2}}\sqrt{10}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}-2\sqrt{10}+\sqrt{18}\right)^{2}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

Combine 2\sqrt{2} and 3\sqrt{2} to get 5\sqrt{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5\sqrt{2}-2\sqrt{10}\right)^{2}.

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$25\times 2-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$50-20\sqrt{2}\sqrt{10}+4\left(\sqrt{10}\right)^{2}$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$50-20\sqrt{2}\sqrt{2}\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

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

$$50-20\times 2\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

Multiply -20 and 2 to get -40.

$$50-40\sqrt{5}+4\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$50-40\sqrt{5}+4\times 10$$

Multiply 4 and 10 to get 40.

$$50-40\sqrt{5}+40$$

Add 50 and 40 to get 90.

$$90-40\sqrt{5}$$