Solve 8m^2-8(m+1)(3m-2)geq0

$$8 m ^ { 2 } - 8 ( m + 1 ) ( 3 m - 2 ) \geq 0$$

Solve for m

$$m\in \begin{bmatrix}\frac{-\sqrt{17}-1}{4},\frac{\sqrt{17}-1}{4}\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply -1 and 8 to get -8.

$$8m^{2}-8\left(m+1\right)\left(3m-2\right)\geq 0$$

Use the distributive property to multiply -8 by m+1.

$$8m^{2}+\left(-8m-8\right)\left(3m-2\right)\geq 0$$

Use the distributive property to multiply -8m-8 by 3m-2 and combine like terms.

$$8m^{2}-24m^{2}-8m+16\geq 0$$

Combine 8m^{2} and -24m^{2} to get -16m^{2}.

$$-16m^{2}-8m+16\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -16m^{2}-8m+16 positive. Since -1 is negative, the inequality direction is changed.

$$16m^{2}+8m-16\leq 0$$

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.

$$16m^{2}+8m-16=0$$

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 16 for a, 8 for b, and -16 for c in the quadratic formula.

$$m=\frac{-8±\sqrt{8^{2}-4\times 16\left(-16\right)}}{2\times 16}$$

Do the calculations.

$$m=\frac{-8±8\sqrt{17}}{32}$$

Solve the equation m=\frac{-8±8\sqrt{17}}{32} when ± is plus and when ± is minus.

$$m=\frac{\sqrt{17}-1}{4} m=\frac{-\sqrt{17}-1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$16\left(m-\frac{\sqrt{17}-1}{4}\right)\left(m-\frac{-\sqrt{17}-1}{4}\right)\leq 0$$

For the product to be ≤0, one of the values m-\frac{\sqrt{17}-1}{4} and m-\frac{-\sqrt{17}-1}{4} has to be ≥0 and the other has to be ≤0. Consider the case when m-\frac{\sqrt{17}-1}{4}\geq 0 and m-\frac{-\sqrt{17}-1}{4}\leq 0.

$$m-\frac{\sqrt{17}-1}{4}\geq 0 m-\frac{-\sqrt{17}-1}{4}\leq 0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m-\frac{\sqrt{17}-1}{4}\leq 0 and m-\frac{-\sqrt{17}-1}{4}\geq 0.

$$m-\frac{-\sqrt{17}-1}{4}\geq 0 m-\frac{\sqrt{17}-1}{4}\leq 0$$

The solution satisfying both inequalities is m\in \left[\frac{-\sqrt{17}-1}{4},\frac{\sqrt{17}-1}{4}\right].

$$m\in \begin{bmatrix}\frac{-\sqrt{17}-1}{4},\frac{\sqrt{17}-1}{4}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$m\in \begin{bmatrix}\frac{-\sqrt{17}-1}{4},\frac{\sqrt{17}-1}{4}\end{bmatrix}$$