Solve -16/sqrt{8+4}

$$\frac { - 16 } { \sqrt { 8 } + 4 }$$

Evaluate

$$4\left(\sqrt{2}-2\right)\approx -2.343145751$$

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}.

$$\frac{-16}{2\sqrt{2}+4}$$

Rationalize the denominator of \frac{-16}{2\sqrt{2}+4} by multiplying numerator and denominator by 2\sqrt{2}-4.

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

Consider \left(2\sqrt{2}+4\right)\left(2\sqrt{2}-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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{-16\left(2\sqrt{2}-4\right)}{8-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

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

Subtract 16 from 8 to get -8.

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

Divide -16\left(2\sqrt{2}-4\right) by -8 to get 2\left(2\sqrt{2}-4\right).

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

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

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