Solved on Dec 19, 2023

Solve the inequality $x^{2} + 5x - 8 \geq x - 3$ and express the solution set in interval notation.

STEP 1

Assumptions
1. We are solving the inequality $x^{2} + 5x - 8 \geq x - 3$ .
2. We will solve for $x$ in the real numbers.
3. The solution will be expressed in interval notation.

STEP 2

First, we need to bring all terms to one side of the inequality to set it to zero.
$x^{2} + 5x - 8 - (x - 3) \geq 0$

STEP 3

Now, simplify the inequality by combining like terms.
$x^{2} + 5x - 8 - x + 3 \geq 0$

STEP 4

Combine the $x$ terms and the constant terms.
$x^{2} + 4x - 5 \geq 0$

STEP 5

Next, we need to factor the quadratic expression if possible.
The factors of -5 that add up to 4 are 5 and -1.

STEP 6

Factor the quadratic expression.
$(x + 5)(x - 1) \geq 0$

STEP 7

Now, we will find the roots of the quadratic by setting each factor equal to zero.
For $x + 5 = 0$ , we have:
$x = -5$
For $x - 1 = 0$ , we have:
$x = 1$

STEP 8

The roots divide the number line into intervals. We will test each interval to see where the inequality holds true.
The intervals are $(-\infty, -5)$ , $(-5, 1)$ , and $(1, \infty)$ .

STEP 9

Choose a test point from each interval and plug it into the factored inequality to determine if the inequality is satisfied.
For $(-\infty, -5)$ , let's test $x = -6$ :
$(-6 + 5)(-6 - 1) \geq 0$ $(-1)(-7) \geq 0$ $7 \geq 0$
This is true, so the interval $(-\infty, -5)$ is part of the solution.

STEP 10

For $(-5, 1)$ , let's test $x = 0$ :
$(0 + 5)(0 - 1) \geq 0$ $(5)(-1) \geq 0$ $-5 \geq 0$
This is false, so the interval $(-5, 1)$ is not part of the solution.

STEP 11

For $(1, \infty)$ , let's test $x = 2$ :
$(2 + 5)(2 - 1) \geq 0$ $(7)(1) \geq 0$ $7 \geq 0$
This is true, so the interval $(1, \infty)$ is part of the solution.

STEP 12

Since the inequality is non-strict ( $\geq$ ), we include the endpoints $-5$ and $1$ in our solution.

STEP 13

Combine the intervals that satisfy the inequality to write the solution in interval notation.
$(-\infty, -5] \cup [1, \infty)$
The solution set is $(-\infty, -5] \cup [1, \infty)$ .

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Solve the inequality $x^{2} + 5x - 8 \geq x - 3$ and express the solution set in interval notation.

STEP 1

Assumptions
1. We are solving the inequality $x^{2} + 5x - 8 \geq x - 3$ .
2. We will solve for $x$ in the real numbers.
3. The solution will be expressed in interval notation.

STEP 2

First, we need to bring all terms to one side of the inequality to set it to zero.
$x^{2} + 5x - 8 - (x - 3) \geq 0$

STEP 3

Now, simplify the inequality by combining like terms.
$x^{2} + 5x - 8 - x + 3 \geq 0$

STEP 4

Combine the $x$ terms and the constant terms.
$x^{2} + 4x - 5 \geq 0$

STEP 5

Next, we need to factor the quadratic expression if possible.
The factors of -5 that add up to 4 are 5 and -1.

STEP 6

Factor the quadratic expression.
$(x + 5)(x - 1) \geq 0$

STEP 7

Now, we will find the roots of the quadratic by setting each factor equal to zero.
For $x + 5 = 0$ , we have:
$x = -5$
For $x - 1 = 0$ , we have:
$x = 1$

STEP 8

The roots divide the number line into intervals. We will test each interval to see where the inequality holds true.
The intervals are $(-\infty, -5)$ , $(-5, 1)$ , and $(1, \infty)$ .

STEP 9

Choose a test point from each interval and plug it into the factored inequality to determine if the inequality is satisfied.
For $(-\infty, -5)$ , let's test $x = -6$ :
$(-6 + 5)(-6 - 1) \geq 0$ $(-1)(-7) \geq 0$ $7 \geq 0$
This is true, so the interval $(-\infty, -5)$ is part of the solution.

STEP 10

For $(-5, 1)$ , let's test $x = 0$ :
$(0 + 5)(0 - 1) \geq 0$ $(5)(-1) \geq 0$ $-5 \geq 0$
This is false, so the interval $(-5, 1)$ is not part of the solution.

STEP 11

For $(1, \infty)$ , let's test $x = 2$ :
$(2 + 5)(2 - 1) \geq 0$ $(7)(1) \geq 0$ $7 \geq 0$
This is true, so the interval $(1, \infty)$ is part of the solution.

STEP 12

Since the inequality is non-strict ( $\geq$ ), we include the endpoints $-5$ and $1$ in our solution.

STEP 13

Combine the intervals that satisfy the inequality to write the solution in interval notation.
$(-\infty, -5] \cup [1, \infty)$
The solution set is $(-\infty, -5] \cup [1, \infty)$ .

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Solve the inequality x2+5x−8≥x−3x^{2} + 5x - 8 \geq x - 3x2+5x−8≥x−3 and express the solution set in interval notation.

STEP 1

Assumptions1. We are solving the inequality x2+5x−8≥x−3 x^{2} + 5x - 8 \geq x - 3 x2+5x−8≥x−3.2. We will solve for x x x in the real numbers.3. The solution will be expressed in interval notation.

STEP 2

First, we need to bring all terms to one side of the inequality to set it to zero.x2+5x−8−(x−3)≥0 x^{2} + 5x - 8 - (x - 3) \geq 0 x2+5x−8−(x−3)≥0

STEP 3

Now, simplify the inequality by combining like terms.x2+5x−8−x+3≥0 x^{2} + 5x - 8 - x + 3 \geq 0 x2+5x−8−x+3≥0

STEP 4

Combine the x x x terms and the constant terms.x2+4x−5≥0 x^{2} + 4x - 5 \geq 0 x2+4x−5≥0

STEP 5

Next, we need to factor the quadratic expression if possible.The factors of -5 that add up to 4 are 5 and -1.

STEP 6

Factor the quadratic expression.(x+5)(x−1)≥0 (x + 5)(x - 1) \geq 0 (x+5)(x−1)≥0

STEP 7

Now, we will find the roots of the quadratic by setting each factor equal to zero.For x+5=0 x + 5 = 0 x+5=0, we have:x=−5 x = -5 x=−5For x−1=0 x - 1 = 0 x−1=0, we have:x=1 x = 1 x=1

STEP 8

The roots divide the number line into intervals. We will test each interval to see where the inequality holds true.The intervals are (−∞,−5) (-\infty, -5) (−∞,−5), (−5,1) (-5, 1) (−5,1), and (1,∞) (1, \infty) (1,∞).

STEP 9

STEP 10

For (−5,1) (-5, 1) (−5,1), let's test x=0 x = 0 x=0:(0+5)(0−1)≥0 (0 + 5)(0 - 1) \geq 0 (0+5)(0−1)≥0 (5)(−1)≥0 (5)(-1) \geq 0 (5)(−1)≥0 −5≥0 -5 \geq 0 −5≥0This is false, so the interval (−5,1) (-5, 1) (−5,1) is not part of the solution.

STEP 11

For (1,∞) (1, \infty) (1,∞), let's test x=2 x = 2 x=2:(2+5)(2−1)≥0 (2 + 5)(2 - 1) \geq 0 (2+5)(2−1)≥0 (7)(1)≥0 (7)(1) \geq 0 (7)(1)≥0 7≥0 7 \geq 0 7≥0This is true, so the interval (1,∞) (1, \infty) (1,∞) is part of the solution.

STEP 12

Since the inequality is non-strict (≥ \geq ≥), we include the endpoints −5 -5 −5 and 1 1 1 in our solution.

STEP 13

Combine the intervals that satisfy the inequality to write the solution in interval notation.(−∞,−5]∪[1,∞) (-\infty, -5] \cup [1, \infty) (−∞,−5]∪[1,∞)The solution set is (−∞,−5]∪[1,∞) (-\infty, -5] \cup [1, \infty) (−∞,−5]∪[1,∞).

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Solve the inequality $x^{2} + 5x - 8 \geq x - 3$ and express the solution set in interval notation.

Assumptions
1. We are solving the inequality $x^{2} + 5x - 8 \geq x - 3$ .
2. We will solve for $x$ in the real numbers.
3. The solution will be expressed in interval notation.

First, we need to bring all terms to one side of the inequality to set it to zero.
$x^{2} + 5x - 8 - (x - 3) \geq 0$

Now, simplify the inequality by combining like terms.
$x^{2} + 5x - 8 - x + 3 \geq 0$

Combine the $x$ terms and the constant terms.
$x^{2} + 4x - 5 \geq 0$

Next, we need to factor the quadratic expression if possible.
The factors of -5 that add up to 4 are 5 and -1.

Factor the quadratic expression.
$(x + 5)(x - 1) \geq 0$

Now, we will find the roots of the quadratic by setting each factor equal to zero.
For $x + 5 = 0$ , we have:
$x = -5$
For $x - 1 = 0$ , we have:
$x = 1$

The roots divide the number line into intervals. We will test each interval to see where the inequality holds true.
The intervals are $(-\infty, -5)$ , $(-5, 1)$ , and $(1, \infty)$ .

For $(-5, 1)$ , let's test $x = 0$ :
$(0 + 5)(0 - 1) \geq 0$ $(5)(-1) \geq 0$ $-5 \geq 0$
This is false, so the interval $(-5, 1)$ is not part of the solution.

For $(1, \infty)$ , let's test $x = 2$ :
$(2 + 5)(2 - 1) \geq 0$ $(7)(1) \geq 0$ $7 \geq 0$
This is true, so the interval $(1, \infty)$ is part of the solution.

Since the inequality is non-strict ( $\geq$ ), we include the endpoints $-5$ and $1$ in our solution.

Combine the intervals that satisfy the inequality to write the solution in interval notation.
$(-\infty, -5] \cup [1, \infty)$
The solution set is $(-\infty, -5] \cup [1, \infty)$ .