Question

Picture attached, solve inequality - need help with interval chart especially

Answer 1

Given:

The expression is:

`(x^2-5x)/(2x^2+4x)\leq(x-8)/(x+6)`

Find-:

Solve the expression

Explanation-:

The expression is:

`(x^2-5x)/(2x^2+4x)\leq(x-8)/(x+6)`

The value of "x" is:

`\begin{gathered} (x^2-5x)/(2(x^2+2x))\leq(-(-x+8))/(x+6) \\ \\ \\ (-x^3+13x^2+2x)/(x(x+2)(x+6))\leq0 \end{gathered}`

So, the "x" is:

The factor is:

`\begin{gathered} (-x(x^2-13x-2))/(x(x+2)(x+6))\leq0 \\ \\ \\ (-x(x-(13+√(177))/(2))(x-(13-√(177))/(2)))/(x(x+2)(x+6))\leq0 \end{gathered}`
`\begin{gathered} (x(x-(13+√(177))/(2))(x-(13-√(177))/(2)))/(x(x+2)(x+6))\ge0 \\ \\ \\ ((x-(13+√(177))/(2))(x-(13-√(177))/(2)))/((x+2)(x+6))\ge0 \end{gathered}`

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"