Solve sqrt{18}-2/sqrt{2}-frac{sqrt{8}}{2}+(sqrt{5}-1)^0

$$\sqrt { 18 } - \frac { 2 } { \sqrt { 2 } } - \frac { \sqrt { 8 } } { 2 } + ( \sqrt { 5 } - 1 ) ^ { 0 }$$

Evaluate

$$\sqrt{2}+1\approx 2.414213562$$

Solution Steps

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}.

$$3\sqrt{2}-\frac{2}{\sqrt{2}}-\frac{\sqrt{8}}{2}+\left(\sqrt{5}-1\right)^{0}$$

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

$$3\sqrt{2}-\frac{2\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{8}}{2}+\left(\sqrt{5}-1\right)^{0}$$

The square of \sqrt{2} is 2.

$$3\sqrt{2}-\frac{2\sqrt{2}}{2}-\frac{\sqrt{8}}{2}+\left(\sqrt{5}-1\right)^{0}$$

Cancel out 2 and 2.

$$3\sqrt{2}-\sqrt{2}-\frac{\sqrt{8}}{2}+\left(\sqrt{5}-1\right)^{0}$$

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}.

$$3\sqrt{2}-\sqrt{2}-\frac{2\sqrt{2}}{2}+\left(\sqrt{5}-1\right)^{0}$$

Cancel out 2 and 2.

$$3\sqrt{2}-\sqrt{2}-\sqrt{2}+\left(\sqrt{5}-1\right)^{0}$$

Combine -\sqrt{2} and -\sqrt{2} to get 2\left(-\sqrt{2}\right).

$$3\sqrt{2}+2\left(-\sqrt{2}\right)+\left(\sqrt{5}-1\right)^{0}$$

Calculate \sqrt{5}-1 to the power of 0 and get 1.

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

Multiply 2 and -1 to get -2.

$$3\sqrt{2}-2\sqrt{2}+1$$

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

$$\sqrt{2}+1$$

Factor

$$\sqrt{2} + 1 = 2.414213562$$