Solve 1/4-(2x+3)1/2+x^2+x-1>0

$$\frac{ 1 }{ 4 } -(2x+3) \frac{ 1 }{ 2 } + { x }^{ 2 } +x-1 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{3}{2}\right)\cup \left(\frac{3}{2},\infty\right)$$

Solution Steps

Use the distributive property to multiply 2x+3 by \frac{1}{2}.

$$\frac{1}{4}-\left(x+\frac{3}{2}\right)+x^{2}+x-1>0$$

To find the opposite of x+\frac{3}{2}, find the opposite of each term.

$$\frac{1}{4}-x-\frac{3}{2}+x^{2}+x-1>0$$

Subtract \frac{3}{2} from \frac{1}{4} to get -\frac{5}{4}.

$$-\frac{5}{4}-x+x^{2}+x-1>0$$

Combine -x and x to get 0.

$$-\frac{5}{4}+x^{2}-1>0$$

Subtract 1 from -\frac{5}{4} to get -\frac{9}{4}.

$$-\frac{9}{4}+x^{2}>0$$

Add \frac{9}{4} to both sides.

$$x^{2}>\frac{9}{4}$$

Calculate the square root of \frac{9}{4} and get \frac{3}{2}. Rewrite \frac{9}{4} as \left(\frac{3}{2}\right)^{2}.

$$x^{2}>\left(\frac{3}{2}\right)^{2}$$

Inequality holds for |x|>\frac{3}{2}.

$$|x|>\frac{3}{2}$$

Rewrite |x|>\frac{3}{2} as x<-\frac{3}{2}\text{; }x>\frac{3}{2}.

$$x<-\frac{3}{2}\text{; }x>\frac{3}{2}$$