Solve (5x-4)^2-(3x+7)^2leq0

$$( 5 x - 4 ) ^ { 2 } - ( 3 x + 7 ) ^ { 2 } \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{3}{8},\frac{11}{2}\end{bmatrix}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(3x+7\right)^{2}\leq 0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+7\right)^{2}.

$$25x^{2}-40x+16-\left(9x^{2}+42x+49\right)\leq 0$$

To find the opposite of 9x^{2}+42x+49, find the opposite of each term.

$$25x^{2}-40x+16-9x^{2}-42x-49\leq 0$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-40x+16-42x-49\leq 0$$

Combine -40x and -42x to get -82x.

$$16x^{2}-82x+16-49\leq 0$$

Subtract 49 from 16 to get -33.

$$16x^{2}-82x-33\leq 0$$

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.

$$16x^{2}-82x-33=0$$

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, -82 for b, and -33 for c in the quadratic formula.

$$x=\frac{-\left(-82\right)±\sqrt{\left(-82\right)^{2}-4\times 16\left(-33\right)}}{2\times 16}$$

Do the calculations.

$$x=\frac{82±94}{32}$$

Solve the equation x=\frac{82±94}{32} when ± is plus and when ± is minus.

$$x=\frac{11}{2} x=-\frac{3}{8}$$

Rewrite the inequality by using the obtained solutions.

$$16\left(x-\frac{11}{2}\right)\left(x+\frac{3}{8}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{11}{2} and x+\frac{3}{8} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{11}{2}\geq 0 and x+\frac{3}{8}\leq 0.

$$x-\frac{11}{2}\geq 0 x+\frac{3}{8}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{11}{2}\leq 0 and x+\frac{3}{8}\geq 0.

$$x+\frac{3}{8}\geq 0 x-\frac{11}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{3}{8},\frac{11}{2}\right].

$$x\in \begin{bmatrix}-\frac{3}{8},\frac{11}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{3}{8},\frac{11}{2}\end{bmatrix}$$