Solve x-4-sqrt{27}geqsqrt{3}(x-x)

$$x - 4 - \sqrt { 27 } \geq \sqrt { 3 } ( x - x )$$

Solve for x

$$x\geq 3\sqrt{3}+4$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$x-4-3\sqrt{3}\geq \sqrt{3}\left(x-x\right)$$

Combine x and -x to get 0.

$$x-4-3\sqrt{3}\geq \sqrt{3}\times 0$$

Anything times zero gives zero.

$$x-4-3\sqrt{3}\geq 0$$

Add 4 to both sides. Anything plus zero gives itself.

$$x-3\sqrt{3}\geq 4$$

Add 3\sqrt{3} to both sides.

$$x\geq 4+3\sqrt{3}$$