Solve 9(x+1)^2-(3x-1)^2geq-4

$$9 ( x + 1 ) ^ { 2 } - ( 3 x - 1 ) ^ { 2 } \geq - 4$$

Solve for x

$$x\geq -\frac{1}{2}$$

Solution Steps

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

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

Use the distributive property to multiply 9 by x^{2}+2x+1.

$$9x^{2}+18x+9-\left(3x-1\right)^{2}\geq -4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}+18x+9-\left(9x^{2}-6x+1\right)\geq -4$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$9x^{2}+18x+9-9x^{2}+6x-1\geq -4$$

Combine 9x^{2} and -9x^{2} to get 0.

$$18x+9+6x-1\geq -4$$

Combine 18x and 6x to get 24x.

$$24x+9-1\geq -4$$

Subtract 1 from 9 to get 8.

$$24x+8\geq -4$$

Subtract 8 from both sides.

$$24x\geq -4-8$$

Subtract 8 from -4 to get -12.

$$24x\geq -12$$

Divide both sides by 24. Since 24 is positive, the inequality direction remains the same.

$$x\geq \frac{-12}{24}$$

Reduce the fraction \frac{-12}{24} to lowest terms by extracting and canceling out 12.

$$x\geq -\frac{1}{2}$$