Solve -8/sqrt{12-4}

$$\frac { - 8 } { \sqrt { 12 } - 4 }$$

Evaluate

$$4\left(\sqrt{3}+2\right)\approx 14.92820323$$

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}.

$$\frac{-8}{2\sqrt{3}-4}$$

Rationalize the denominator of \frac{-8}{2\sqrt{3}-4} by multiplying numerator and denominator by 2\sqrt{3}+4.

$$\frac{-8\left(2\sqrt{3}+4\right)}{\left(2\sqrt{3}-4\right)\left(2\sqrt{3}+4\right)}$$

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

$$\frac{-8\left(2\sqrt{3}+4\right)}{\left(2\sqrt{3}\right)^{2}-4^{2}}$$

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

$$\frac{-8\left(2\sqrt{3}+4\right)}{2^{2}\left(\sqrt{3}\right)^{2}-4^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{-8\left(2\sqrt{3}+4\right)}{4\left(\sqrt{3}\right)^{2}-4^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{-8\left(2\sqrt{3}+4\right)}{4\times 3-4^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{-8\left(2\sqrt{3}+4\right)}{12-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{-8\left(2\sqrt{3}+4\right)}{12-16}$$

Subtract 16 from 12 to get -4.

$$\frac{-8\left(2\sqrt{3}+4\right)}{-4}$$

Divide -8\left(2\sqrt{3}+4\right) by -4 to get 2\left(2\sqrt{3}+4\right).

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

Use the distributive property to multiply 2 by 2\sqrt{3}+4.

$$4\sqrt{3}+8$$