Solve 1/4+sqrt{12}

$$\frac { 1 } { 4 + \sqrt { 12 } }$$

Evaluate

$$-\frac{\sqrt{3}}{2}+1\approx 0.133974596$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{1}{4+2\sqrt{3}}$$

Rationalize the denominator of \frac{1}{4+2\sqrt{3}} by multiplying numerator and denominator by 4-2\sqrt{3}.

$$\frac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}$$

Consider \left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4-2\sqrt{3}}{4^{2}-\left(2\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{4-2\sqrt{3}}{16-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{4-2\sqrt{3}}{16-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4-2\sqrt{3}}{16-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{4-2\sqrt{3}}{16-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{4-2\sqrt{3}}{16-12}$$

Subtract 12 from 16 to get 4.

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