Solve (sqrt{a}+sqrt{8a})^2=54+bsqrt{2}

$$( \sqrt { a } + \sqrt { 8 a } ) ^ { 2 } = 54 + b \sqrt { 2 }$$

Solve for b

$$b=\frac{9\sqrt{2}a}{2}+4a-27\sqrt{2}$$

$$a\geq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\sqrt{a}+\sqrt{8a}\right)^{2}.

$$\left(\sqrt{a}\right)^{2}+2\sqrt{a}\sqrt{8a}+\left(\sqrt{8a}\right)^{2}=54+b\sqrt{2}$$

Calculate \sqrt{a} to the power of 2 and get a.

$$a+2\sqrt{a}\sqrt{8a}+\left(\sqrt{8a}\right)^{2}=54+b\sqrt{2}$$

Calculate \sqrt{8a} to the power of 2 and get 8a.

$$a+2\sqrt{a}\sqrt{8a}+8a=54+b\sqrt{2}$$

Combine a and 8a to get 9a.

$$9a+2\sqrt{a}\sqrt{8a}=54+b\sqrt{2}$$

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

$$54+b\sqrt{2}=9a+2\sqrt{a}\sqrt{8a}$$

Subtract 54 from both sides.

$$b\sqrt{2}=9a+2\sqrt{a}\sqrt{8a}-54$$

The equation is in standard form.

$$\sqrt{2}b=2\sqrt{a}\sqrt{8a}+9a-54$$

Divide both sides by \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{4\sqrt{2}a+9a-54}{\sqrt{2}}$$

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

$$b=\frac{4\sqrt{2}a+9a-54}{\sqrt{2}}$$

Divide 9a+4a\sqrt{2}-54 by \sqrt{2}.

$$b=\frac{\sqrt{2}\left(4\sqrt{2}a+9a-54\right)}{2}$$

Solve for a

$$a=-\frac{\left(4\sqrt{2}-9\right)\left(\sqrt{2}b+54\right)}{49}$$

$$b\geq -27\sqrt{2}$$