QuestionFind the range of $x$ for the inequality $-6 < x^{2} - 5x \leq 0$ and show the solution on a number line.

Studdy Solution

STEP 1

Assumptions1. The inequality to solve is $-6<x^{}-5x\leq0$ . . The solution should be illustrated on a number line.

STEP 2

We can rewrite the inequality as $x^{2}-5x+6>0$ .

STEP 3

We can factor the quadratic expression on the left side of the inequality.
$x^{2}-5x+6 = (x-2)(x-3)$

STEP 4

Now we have the inequality $(x-2)(x-3)>0$ .

STEP 5

The inequality $(x-2)(x-3)>0$ will hold true when both factors are positive or both are negative.

STEP 6

We set each factor equal to zero and solve for $x$ to find the critical points.
$x-2=0 \Rightarrow x=2$ $x-3=0 \Rightarrow x=3$

STEP 7

We now have three intervals to consider $(-\infty,2)$ , $(2,3)$ , and $(3, \infty)$ .

STEP 8

We choose test points in each interval and substitute them into the inequality to determine if the inequality holds true.For $(-\infty,2)$ , we choose $x=1$ . For $(2,3)$ , we choose $x=2.5$ . For $(3, \infty)$ , we choose $x=4$ .

STEP 9

Substitute $x=$ into the inequality.
$(-2)(-3)>$ $(-)(-2)>$ $2>$ The inequality holds true for the interval $(-\infty,2)$ .

STEP 10

Substitute $x=2.5$ into the inequality.
$(2.5-2)(2.5-3)>0$ $(0.5)(-0.5)>0$ $-0.25>0$ The inequality does not hold true for the interval $(2,3)$ .

STEP 11

Substitute $x=4$ into the inequality.
$(4-)(4-3)>0$ $()()>0$ $>0$ The inequality holds true for the interval $(3, \infty)$ .

STEP 12

From the above steps, we can conclude that the solution to the inequality is $x \in (-\infty,2) \cup (, \infty)$ .

STEP 13

On a number line, we can represent this solution as an open circle at $x=2$ and $x=3$ , with arrows pointing to the left of $x=2$ and to the right of $x=3$ .
The range of values of $x$ which satisfies the inequality $-6<x^{2}-5x\leq0$ is $x \in (-\infty,2) \cup (3, \infty)$ .

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