Solve frac{sqrt{6}}{1+sqrt{24}}

$$\frac { \sqrt { 6 } } { 1 + \sqrt { 24 } }$$

Evaluate

$$\frac{12-\sqrt{6}}{23}\approx 0.415239576$$

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}.

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

Rationalize the denominator of \frac{\sqrt{6}}{1+2\sqrt{6}} by multiplying numerator and denominator by 1-2\sqrt{6}.

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

Consider \left(1+2\sqrt{6}\right)\left(1-2\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 1 to the power of 2 and get 1.

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

Expand \left(2\sqrt{6}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{6}\left(1-2\sqrt{6}\right)}{1-4\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{\sqrt{6}\left(1-2\sqrt{6}\right)}{1-4\times 6}$$

Multiply 4 and 6 to get 24.

$$\frac{\sqrt{6}\left(1-2\sqrt{6}\right)}{1-24}$$

Subtract 24 from 1 to get -23.

$$\frac{\sqrt{6}\left(1-2\sqrt{6}\right)}{-23}$$

Use the distributive property to multiply \sqrt{6} by 1-2\sqrt{6}.

$$\frac{\sqrt{6}-2\left(\sqrt{6}\right)^{2}}{-23}$$

The square of \sqrt{6} is 6.

$$\frac{\sqrt{6}-2\times 6}{-23}$$

Multiply -2 and 6 to get -12.

$$\frac{\sqrt{6}-12}{-23}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{6}+12}{23}$$