Solve (x+2)^2+3(x+1)^2<(2x+1)^2+7

$$( x + 2 ) ^ { 2 } + 3 ( x + 1 ) ^ { 2 } < ( 2 x + 1 ) ^ { 2 } + 7$$

Solve for x

$$x<\frac{1}{6}$$

Solution Steps

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

$$x^{2}+4x+4+3\left(x+1\right)^{2}<\left(2x+1\right)^{2}+7$$

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

$$x^{2}+4x+4+3\left(x^{2}+2x+1\right)<\left(2x+1\right)^{2}+7$$

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

$$x^{2}+4x+4+3x^{2}+6x+3<\left(2x+1\right)^{2}+7$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}+4x+4+6x+3<\left(2x+1\right)^{2}+7$$

Combine 4x and 6x to get 10x.

$$4x^{2}+10x+4+3<\left(2x+1\right)^{2}+7$$

Add 4 and 3 to get 7.

$$4x^{2}+10x+7<\left(2x+1\right)^{2}+7$$

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

$$4x^{2}+10x+7<4x^{2}+4x+1+7$$

Add 1 and 7 to get 8.

$$4x^{2}+10x+7<4x^{2}+4x+8$$

Subtract 4x^{2} from both sides.

$$4x^{2}+10x+7-4x^{2}<4x+8$$

Combine 4x^{2} and -4x^{2} to get 0.

$$10x+7<4x+8$$

Subtract 4x from both sides.

$$10x+7-4x<8$$

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

$$6x+7<8$$

Subtract 7 from both sides.

$$6x<8-7$$

Subtract 7 from 8 to get 1.

$$6x<1$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

$$x<\frac{1}{6}$$