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

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

Solve for x

$$x>\frac{3}{8}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{3}{2}\right)^{2}.

$$x^{2}-3x+\frac{9}{4}+2x\left(x-\frac{1}{2}\right)<3\left(x^{2}+\frac{1}{4}\right)$$

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

$$x^{2}-3x+\frac{9}{4}+2x^{2}-x<3\left(x^{2}+\frac{1}{4}\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-3x+\frac{9}{4}-x<3\left(x^{2}+\frac{1}{4}\right)$$

Combine -3x and -x to get -4x.

$$3x^{2}-4x+\frac{9}{4}<3\left(x^{2}+\frac{1}{4}\right)$$

Use the distributive property to multiply 3 by x^{2}+\frac{1}{4}.

$$3x^{2}-4x+\frac{9}{4}<3x^{2}+\frac{3}{4}$$

Subtract 3x^{2} from both sides.

$$3x^{2}-4x+\frac{9}{4}-3x^{2}<\frac{3}{4}$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-4x+\frac{9}{4}<\frac{3}{4}$$

Subtract \frac{9}{4} from both sides.

$$-4x<\frac{3}{4}-\frac{9}{4}$$

Subtract \frac{9}{4} from \frac{3}{4} to get -\frac{3}{2}.

$$-4x<-\frac{3}{2}$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{-\frac{3}{2}}{-4}$$

Express \frac{-\frac{3}{2}}{-4} as a single fraction.

$$x>\frac{-3}{2\left(-4\right)}$$

Multiply 2 and -4 to get -8.

$$x>\frac{-3}{-8}$$

Fraction \frac{-3}{-8} can be simplified to \frac{3}{8} by removing the negative sign from both the numerator and the denominator.

$$x>\frac{3}{8}$$