Solve x+sqrt{x}<(2sqrt{x}+3)(sqrt{x}-1)

$$x + \sqrt { x } < ( 2 \sqrt { x } + 3 ) ( \sqrt { x } - 1 )$$

Solve for x

$$x>3$$

Solution Steps

Apply the distributive property by multiplying each term of 2\sqrt{x}+3 by each term of \sqrt{x}-1.

$$x+\sqrt{x}<2\left(\sqrt{x}\right)^{2}-2\sqrt{x}+3\sqrt{x}-3$$

Calculate \sqrt{x} to the power of 2 and get x.

$$x+\sqrt{x}<2x-2\sqrt{x}+3\sqrt{x}-3$$

Combine -2\sqrt{x} and 3\sqrt{x} to get \sqrt{x}.

$$x+\sqrt{x}<2x+\sqrt{x}-3$$

Subtract 2x from both sides.

$$x+\sqrt{x}-2x<\sqrt{x}-3$$

Combine x and -2x to get -x.

$$-x+\sqrt{x}<\sqrt{x}-3$$

Subtract \sqrt{x} from both sides.

$$-x+\sqrt{x}-\sqrt{x}<-3$$

Combine \sqrt{x} and -\sqrt{x} to get 0.

$$-x<-3$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x>\frac{-3}{-1}$$

Fraction \frac{-3}{-1} can be simplified to 3 by removing the negative sign from both the numerator and the denominator.

$$x>3$$