Solve 2-(1+x)(1+x)<x(2-x)

$$2 - ( 1 + x ) ( 1 + x ) < x ( 2 - x )$$

Solve for x

$$x>\frac{1}{4}$$

Solution Steps

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$2-\left(1+x\right)^{2}<x\left(2-x\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$2-\left(1+2x+x^{2}\right)<x\left(2-x\right)$$

To find the opposite of 1+2x+x^{2}, find the opposite of each term.

$$2-1-2x-x^{2}<x\left(2-x\right)$$

Subtract 1 from 2 to get 1.

$$1-2x-x^{2}<x\left(2-x\right)$$

Use the distributive property to multiply x by 2-x.

$$1-2x-x^{2}<2x-x^{2}$$

Subtract 2x from both sides.

$$1-2x-x^{2}-2x<-x^{2}$$

Combine -2x and -2x to get -4x.

$$1-4x-x^{2}<-x^{2}$$

Add x^{2} to both sides.

$$1-4x-x^{2}+x^{2}<0$$

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

$$1-4x<0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-4x<-1$$

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

$$x>\frac{-1}{-4}$$

Fraction \frac{-1}{-4} can be simplified to \frac{1}{4} by removing the negative sign from both the numerator and the denominator.

$$x>\frac{1}{4}$$