Question

In Problems 5–16, use a graphing utility to approximate the real solutions, if any, of each equation rounded to two decimal places. All solutions lie between - 10 and 10.$-\frac{2}{3}{x}^4-2x^3+\frac{5}{2}x=-\frac{2}{3}{x}^2+12$

Short answer

There is no value of X which satisfies this equation that is why$\mathrm{No}\hspace{0.17em}\mathrm{Solution}\hspace{0.17em}\mathrm{for}\hspace{0.17em}x\in \mathrm{ℝ}$

Step-by-step solution

Step 1. Given equation and simplification

$$\begin{gathered} -\frac{2}{3}{x}^4-2x^3+\frac{5}{2}x=-\frac{2}{3}{x}^2+12 \\ -\frac{2}{3}{x}^4·\hspace{0.17em}6-2x^3·\hspace{0.17em}6+\frac{5}{2}x·\hspace{0.17em}6=-\frac{2}{3}{x}^2·\hspace{0.17em}6+12·\hspace{0.17em}6 \\ -4x^4-12x^3+15x=-4x^2+72 \\ -4x^4-12x^3+15x+4x^2-72=0 \end{gathered}$$

Step 2. Graph and X-intersection

No X- intercepts so here no real solution for this equation.