Solve left(1-2xright)^2+left(x-3right)^2<2x-x(3-5x)+1

$${ \left(1-2x \right) }^{ 2 } + { \left(x-3 \right) }^{ 2 } < 2x-x(3-5x)+1$$

Solve for x

$$x>1$$

Solution Steps

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

$$1-4x+4x^{2}+\left(x-3\right)^{2}<2x-x\left(3-5x\right)+1$$

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

$$1-4x+4x^{2}+x^{2}-6x+9<2x-x\left(3-5x\right)+1$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$1-4x+5x^{2}-6x+9<2x-x\left(3-5x\right)+1$$

Combine -4x and -6x to get -10x.

$$1-10x+5x^{2}+9<2x-x\left(3-5x\right)+1$$

Add 1 and 9 to get 10.

$$10-10x+5x^{2}<2x-x\left(3-5x\right)+1$$

Use the distributive property to multiply x by 3-5x.

$$10-10x+5x^{2}<2x-\left(3x-5x^{2}\right)+1$$

To find the opposite of 3x-5x^{2}, find the opposite of each term.

$$10-10x+5x^{2}<2x-3x+5x^{2}+1$$

Combine 2x and -3x to get -x.

$$10-10x+5x^{2}<-x+5x^{2}+1$$

Add x to both sides.

$$10-10x+5x^{2}+x<5x^{2}+1$$

Combine -10x and x to get -9x.

$$10-9x+5x^{2}<5x^{2}+1$$

Subtract 5x^{2} from both sides.

$$10-9x+5x^{2}-5x^{2}<1$$

Combine 5x^{2} and -5x^{2} to get 0.

$$10-9x<1$$

Subtract 10 from both sides.

$$-9x<1-10$$

Subtract 10 from 1 to get -9.

$$-9x<-9$$

Divide both sides by -9. Since -9 is negative, the inequality direction is changed.

$$x>\frac{-9}{-9}$$

Divide -9 by -9 to get 1.

$$x>1$$