Solve (3sqrt{2}+2)^2+(6-sqrt{2})^2

$$( 3 \sqrt { 2 } + 2 ) ^ { 2 } + ( 6 - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$60$$

Solution Steps

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

$$9\left(\sqrt{2}\right)^{2}+12\sqrt{2}+4+\left(6-\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2+12\sqrt{2}+4+\left(6-\sqrt{2}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18+12\sqrt{2}+4+\left(6-\sqrt{2}\right)^{2}$$

Add 18 and 4 to get 22.

$$22+12\sqrt{2}+\left(6-\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{2}\right)^{2}.

$$22+12\sqrt{2}+36-12\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$22+12\sqrt{2}+36-12\sqrt{2}+2$$

Add 36 and 2 to get 38.

$$22+12\sqrt{2}+38-12\sqrt{2}$$

Add 22 and 38 to get 60.

$$60+12\sqrt{2}-12\sqrt{2}$$

Combine 12\sqrt{2} and -12\sqrt{2} to get 0.

$$60$$

Factor

$$2^{2}\times 3\times 5$$