Solve (3+sqrt{8})^2

$$( 3 + \sqrt { 8 } ) ^ { 2 }$$

Evaluate

$$12\sqrt{2}+17\approx 33.970562748$$

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}.

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

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

$$9+12\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+12\sqrt{2}+4\times 2$$

Multiply 4 and 2 to get 8.

$$9+12\sqrt{2}+8$$

Add 9 and 8 to get 17.

$$17+12\sqrt{2}$$

Expand

$$12 \sqrt{2} + 17 = 33.970562748$$

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}.

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

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

$$9+12\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+12\sqrt{2}+4\times 2$$

Multiply 4 and 2 to get 8.

$$9+12\sqrt{2}+8$$

Add 9 and 8 to get 17.

$$17+12\sqrt{2}$$