Solve 3x(2x+3)<2x(x+4.5)+2

$$3 x ( 2 x + 3 ) < 2 x ( x + 4.5 ) + 2$$

Solve for x

$$x\in \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$$

Solution Steps

Use the distributive property to multiply 3x by 2x+3.

$$6x^{2}+9x<2x\left(x+4.5\right)+2$$

Use the distributive property to multiply 2x by x+4.5.

$$6x^{2}+9x<2x^{2}+9x+2$$

Subtract 2x^{2} from both sides.

$$6x^{2}+9x-2x^{2}<9x+2$$

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

$$4x^{2}+9x<9x+2$$

Subtract 9x from both sides.

$$4x^{2}+9x-9x<2$$

Combine 9x and -9x to get 0.

$$4x^{2}<2$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x^{2}<\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}<\frac{1}{2}$$

Calculate the square root of \frac{1}{2} and get \frac{\sqrt{2}}{2}. Rewrite \frac{1}{2} as \left(\frac{\sqrt{2}}{2}\right)^{2}.

$$x^{2}<\left(\frac{\sqrt{2}}{2}\right)^{2}$$

Inequality holds for |x|<\frac{\sqrt{2}}{2}.

$$|x|<\frac{\sqrt{2}}{2}$$

Rewrite |x|<\frac{\sqrt{2}}{2} as x\in \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).

$$x\in \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$$