Given:
The expression is:
[tex]\frac{x^2-5x}{2x^2+4x}\leq\frac{x-8}{x+6}[/tex]
Find-:
Solve the expression
Explanation-:
The expression is:
[tex]\frac{x^2-5x}{2x^2+4x}\leq\frac{x-8}{x+6}[/tex]
The value of "x" is:
[tex]\begin{gathered} \frac{x^2-5x}{2(x^2+2x)}\leq\frac{-(-x+8)}{x+6} \\ \\ \\ \frac{-x^3+13x^2+2x}{x(x+2)(x+6)}\leq0 \end{gathered}[/tex]
So, the "x" is:
The factor is:
[tex]\begin{gathered} \frac{-x(x^2-13x-2)}{x(x+2)(x+6)}\leq0 \\ \\ \\ \frac{-x(x-\frac{13+\sqrt{177}}{2})(x-\frac{13-\sqrt{177}}{2})}{x(x+2)(x+6)}\leq0 \end{gathered}[/tex][tex]\begin{gathered} \frac{x(x-\frac{13+\sqrt{177}}{2})(x-\frac{13-\sqrt{177}}{2})}{x(x+2)(x+6)}\ge0 \\ \\ \\ \frac{(x-\frac{13+\sqrt{177}}{2})(x-\frac{13-\sqrt{177}}{2})}{(x+2)(x+6)}\ge0 \end{gathered}[/tex]
The "x" in interval is:
[tex]\begin{gathered} \frac{(x-\frac{13+\sqrt{177}}{2})(x-\frac{13-\sqrt{177}}{2})}{(x+2)(x+6)}\ge0 \\ \\ x<-6\text{ or }-2Is the value for "x"
