Solve (sqrt{82}-3sqrt{2})^2

$$( \sqrt { 82 } - 3 \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$100-12\sqrt{41}\approx 23.162509151$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{82}-3\sqrt{2}\right)^{2}.

$$\left(\sqrt{82}\right)^{2}-6\sqrt{82}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{82} is 82.

$$82-6\sqrt{82}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Factor 82=2\times 41. Rewrite the square root of the product \sqrt{2\times 41} as the product of square roots \sqrt{2}\sqrt{41}.

$$82-6\sqrt{2}\sqrt{41}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$82-6\times 2\sqrt{41}+9\left(\sqrt{2}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$82-12\sqrt{41}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$82-12\sqrt{41}+9\times 2$$

Multiply 9 and 2 to get 18.

$$82-12\sqrt{41}+18$$

Add 82 and 18 to get 100.

$$100-12\sqrt{41}$$

Expand

$$100-12\sqrt{41}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{82}-3\sqrt{2}\right)^{2}.

$$\left(\sqrt{82}\right)^{2}-6\sqrt{82}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{82} is 82.

$$82-6\sqrt{82}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Factor 82=2\times 41. Rewrite the square root of the product \sqrt{2\times 41} as the product of square roots \sqrt{2}\sqrt{41}.

$$82-6\sqrt{2}\sqrt{41}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$82-6\times 2\sqrt{41}+9\left(\sqrt{2}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$82-12\sqrt{41}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$82-12\sqrt{41}+9\times 2$$

Multiply 9 and 2 to get 18.

$$82-12\sqrt{41}+18$$

Add 82 and 18 to get 100.

$$100-12\sqrt{41}$$