Solve (1+sqrt{2})^4=a+bsqrt{2}

$$( 1 + \sqrt { 2 } ) ^ { 4 } = a + b \sqrt { 2 }$$

Solve for a

$$a=\sqrt{2}\left(12-b\right)+17$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Subtract b\sqrt{2} from both sides.

$$a=\left(1+\sqrt{2}\right)^{4}-b\sqrt{2}$$

Reorder the terms.

$$a=-\sqrt{2}b+\left(\sqrt{2}+1\right)^{4}$$

Solve for b

$$b=-\frac{\sqrt{2}\left(a-12\sqrt{2}-17\right)}{2}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Subtract a from both sides.

$$b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}-a$$

The equation is in standard form.

$$\sqrt{2}b=-a+\left(\sqrt{2}+1\right)^{4}$$

Divide both sides by \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

Dividing by \sqrt{2} undoes the multiplication by \sqrt{2}.

$$b=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

Divide 17+12\sqrt{2}-a by \sqrt{2}.

$$b=\frac{\sqrt{2}\left(-a+12\sqrt{2}+17\right)}{2}$$