Solve left(3x+1right)^2geq8x(x+1)

$${ \left(3x+1 \right) }^{ 2 } \geq 8x(x+1)$$

Solve for x

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

Steps Using the Quadratic Formula

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

$$9x^{2}+6x+1\geq 8x\left(x+1\right)$$

Use the distributive property to multiply 8x by x+1.

$$9x^{2}+6x+1\geq 8x^{2}+8x$$

Subtract 8x^{2} from both sides.

$$9x^{2}+6x+1-8x^{2}\geq 8x$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{2}+6x+1\geq 8x$$

Subtract 8x from both sides.

$$x^{2}+6x+1-8x\geq 0$$

Combine 6x and -8x to get -2x.

$$x^{2}-2x+1\geq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-2x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{2±0}{2}$$

Solutions are the same.

$$x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-1\right)^{2}\geq 0$$

Inequality holds for x\in \mathrm{R}.

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