Solve frac{sqrt{24}+sqrt{8}}{sqrt{2}}-1

$$\frac{ \sqrt{ 24 } + \sqrt{ 8 } }{ \sqrt{ 2 } } -1$$

Evaluate

$$2\sqrt{3}+1\approx 4.464101615$$

Short Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\frac{2\sqrt{6}+\sqrt{8}}{\sqrt{2}}-1$$

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}.

$$\frac{2\sqrt{6}+2\sqrt{2}}{\sqrt{2}}-1$$

Rationalize the denominator of \frac{2\sqrt{6}+2\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-1$$

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

$$\frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\frac{2}{2}$$

Since \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}-2}{2}$$

Do the multiplications in \left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}-2.

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

Do the calculations in 4\sqrt{3}+4-2.

$$\frac{4\sqrt{3}+2}{2}$$

Divide each term of 4\sqrt{3}+2 by 2 to get 2\sqrt{3}+1.

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