Solve (x-1)^2+(x+3)^2leq2x(x+2)+10

$$( x - 1 ) ^ { 2 } + ( x + 3 ) ^ { 2 } \leq 2 x ( x + 2 ) + 10$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

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

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

$$x^{2}-2x+1+x^{2}+6x+9\leq 2x\left(x+2\right)+10$$

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

$$2x^{2}-2x+1+6x+9\leq 2x\left(x+2\right)+10$$

Combine -2x and 6x to get 4x.

$$2x^{2}+4x+1+9\leq 2x\left(x+2\right)+10$$

Add 1 and 9 to get 10.

$$2x^{2}+4x+10\leq 2x\left(x+2\right)+10$$

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

$$2x^{2}+4x+10\leq 2x^{2}+4x+10$$

Subtract 2x^{2} from both sides.

$$2x^{2}+4x+10-2x^{2}\leq 4x+10$$

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

$$4x+10\leq 4x+10$$

Subtract 4x from both sides.

$$4x+10-4x\leq 10$$

Combine 4x and -4x to get 0.

$$10\leq 10$$

This is true for any x.

$$x\in \mathrm{R}$$