Solve frac{sqrt{6}}{5-2sqrt{8}}-sqrt{150}

$$\frac { \sqrt { 6 } } { 5 - 2 \sqrt { 8 } } - \sqrt { 150 }$$

Evaluate

$$\frac{-8\sqrt{3}-40\sqrt{6}}{7}\approx -15.976570882$$

Short 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}.

$$\frac{\sqrt{6}}{5-2\times 2\sqrt{2}}-\sqrt{150}$$

Multiply -2 and 2 to get -4.

$$\frac{\sqrt{6}}{5-4\sqrt{2}}-\sqrt{150}$$

Rationalize the denominator of \frac{\sqrt{6}}{5-4\sqrt{2}} by multiplying numerator and denominator by 5+4\sqrt{2}.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{\left(5-4\sqrt{2}\right)\left(5+4\sqrt{2}\right)}-\sqrt{150}$$

Consider \left(5-4\sqrt{2}\right)\left(5+4\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{5^{2}-\left(-4\sqrt{2}\right)^{2}}-\sqrt{150}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{25-\left(-4\sqrt{2}\right)^{2}}-\sqrt{150}$$

Expand \left(-4\sqrt{2}\right)^{2}.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{25-\left(-4\right)^{2}\left(\sqrt{2}\right)^{2}}-\sqrt{150}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{25-16\left(\sqrt{2}\right)^{2}}-\sqrt{150}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{25-16\times 2}-\sqrt{150}$$

Multiply 16 and 2 to get 32.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{25-32}-\sqrt{150}$$

Subtract 32 from 25 to get -7.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{-7}-\sqrt{150}$$

Factor 150=5^{2}\times 6. Rewrite the square root of the product \sqrt{5^{2}\times 6} as the product of square roots \sqrt{5^{2}}\sqrt{6}. Take the square root of 5^{2}.

$$\frac{\sqrt{6}\left(5+4\sqrt{2}\right)}{-7}-5\sqrt{6}$$

Use the distributive property to multiply \sqrt{6} by 5+4\sqrt{2}.

$$\frac{5\sqrt{6}+4\sqrt{6}\sqrt{2}}{-7}-5\sqrt{6}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{5\sqrt{6}+4\sqrt{2}\sqrt{3}\sqrt{2}}{-7}-5\sqrt{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{5\sqrt{6}+4\times 2\sqrt{3}}{-7}-5\sqrt{6}$$

Multiply 4 and 2 to get 8.

$$\frac{5\sqrt{6}+8\sqrt{3}}{-7}-5\sqrt{6}$$

Multiply both numerator and denominator by -1.

$$\frac{-5\sqrt{6}-8\sqrt{3}}{7}-5\sqrt{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -5\sqrt{6} times \frac{7}{7}.

$$\frac{-5\sqrt{6}-8\sqrt{3}}{7}+\frac{7\left(-5\right)\sqrt{6}}{7}$$

Since \frac{-5\sqrt{6}-8\sqrt{3}}{7} and \frac{7\left(-5\right)\sqrt{6}}{7} have the same denominator, add them by adding their numerators.

$$\frac{-5\sqrt{6}-8\sqrt{3}+7\left(-5\right)\sqrt{6}}{7}$$

Do the multiplications in -5\sqrt{6}-8\sqrt{3}+7\left(-5\right)\sqrt{6}.

$$\frac{-5\sqrt{6}-8\sqrt{3}-35\sqrt{6}}{7}$$

Do the calculations in -5\sqrt{6}-8\sqrt{3}-35\sqrt{6}.

$$\frac{-40\sqrt{6}-8\sqrt{3}}{7}$$