Solve sqrt{8}+sqrt{18}+sqrt{1/8}=a+bsqrt{2}

$$\sqrt { 8 } + \sqrt { 18 } + \sqrt { \frac { 1 } { 8 } } = a + b \sqrt { 2 }$$

Solve for a

$$a=\frac{\sqrt{2}\left(21-4b\right)}{4}$$

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}.

$$2\sqrt{2}+\sqrt{18}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2\sqrt{2}+3\sqrt{2}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Combine 2\sqrt{2} and 3\sqrt{2} to get 5\sqrt{2}.

$$5\sqrt{2}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$5\sqrt{2}+\frac{\sqrt{1}}{\sqrt{8}}=a+b\sqrt{2}$$

Calculate the square root of 1 and get 1.

$$5\sqrt{2}+\frac{1}{\sqrt{8}}=a+b\sqrt{2}$$

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}.

$$5\sqrt{2}+\frac{1}{2\sqrt{2}}=a+b\sqrt{2}$$

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$5\sqrt{2}+\frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}=a+b\sqrt{2}$$

The square of \sqrt{2} is 2.

$$5\sqrt{2}+\frac{\sqrt{2}}{2\times 2}=a+b\sqrt{2}$$

Multiply 2 and 2 to get 4.

$$5\sqrt{2}+\frac{\sqrt{2}}{4}=a+b\sqrt{2}$$

Combine 5\sqrt{2} and \frac{\sqrt{2}}{4} to get \frac{21}{4}\sqrt{2}.

$$\frac{21}{4}\sqrt{2}=a+b\sqrt{2}$$

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

$$a+b\sqrt{2}=\frac{21}{4}\sqrt{2}$$

Subtract b\sqrt{2} from both sides.

$$a=\frac{21}{4}\sqrt{2}-b\sqrt{2}$$

Reorder the terms.

$$a=-\sqrt{2}b+\frac{21}{4}\sqrt{2}$$

Solve for b

$$b=-\frac{\sqrt{2}a}{2}+\frac{21}{4}$$

Steps for Solving Linear Equation

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}.

$$2\sqrt{2}+\sqrt{18}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2\sqrt{2}+3\sqrt{2}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Combine 2\sqrt{2} and 3\sqrt{2} to get 5\sqrt{2}.

$$5\sqrt{2}+\sqrt{\frac{1}{8}}=a+b\sqrt{2}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$5\sqrt{2}+\frac{\sqrt{1}}{\sqrt{8}}=a+b\sqrt{2}$$

Calculate the square root of 1 and get 1.

$$5\sqrt{2}+\frac{1}{\sqrt{8}}=a+b\sqrt{2}$$

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}.

$$5\sqrt{2}+\frac{1}{2\sqrt{2}}=a+b\sqrt{2}$$

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$5\sqrt{2}+\frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}=a+b\sqrt{2}$$

The square of \sqrt{2} is 2.

$$5\sqrt{2}+\frac{\sqrt{2}}{2\times 2}=a+b\sqrt{2}$$

Multiply 2 and 2 to get 4.

$$5\sqrt{2}+\frac{\sqrt{2}}{4}=a+b\sqrt{2}$$

Combine 5\sqrt{2} and \frac{\sqrt{2}}{4} to get \frac{21}{4}\sqrt{2}.

$$\frac{21}{4}\sqrt{2}=a+b\sqrt{2}$$

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

$$a+b\sqrt{2}=\frac{21}{4}\sqrt{2}$$

Subtract a from both sides.

$$b\sqrt{2}=\frac{21}{4}\sqrt{2}-a$$

The equation is in standard form.

$$\sqrt{2}b=-a+\frac{21\sqrt{2}}{4}$$

Divide both sides by \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+\frac{21\sqrt{2}}{4}}{\sqrt{2}}$$

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

$$b=\frac{-a+\frac{21\sqrt{2}}{4}}{\sqrt{2}}$$

Divide \frac{21\sqrt{2}}{4}-a by \sqrt{2}.

$$b=-\frac{\sqrt{2}a}{2}+\frac{21}{4}$$