Solve F(1+sqrt{2})=2(1+sqrt{2})^2+3(1+sqrt{2})-4

$$F ( 1 + \sqrt { 2 } ) = 2 ( 1 + \sqrt { 2 } ) ^ { 2 } + 3 ( 1 + \sqrt { 2 } ) - 4$$

Evaluate

$$7\sqrt{2}+5\approx 14.899494937$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{2}\right)^{2}.

$$2\left(1+2\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)+3\left(1+\sqrt{2}\right)-4$$

The square of \sqrt{2} is 2.

$$2\left(1+2\sqrt{2}+2\right)+3\left(1+\sqrt{2}\right)-4$$

Add 1 and 2 to get 3.

$$2\left(3+2\sqrt{2}\right)+3\left(1+\sqrt{2}\right)-4$$

Use the distributive property to multiply 2 by 3+2\sqrt{2}.

$$6+4\sqrt{2}+3\left(1+\sqrt{2}\right)-4$$

Use the distributive property to multiply 3 by 1+\sqrt{2}.

$$6+4\sqrt{2}+3+3\sqrt{2}-4$$

Add 6 and 3 to get 9.

$$9+4\sqrt{2}+3\sqrt{2}-4$$

Combine 4\sqrt{2} and 3\sqrt{2} to get 7\sqrt{2}.

$$9+7\sqrt{2}-4$$

Subtract 4 from 9 to get 5.

$$5+7\sqrt{2}$$