Algebra Examples

2 x^{2} + 5 x - 3 < 0

$2 x^{2} + 5 x - 3 < 0$

Step 1

Convert the inequality to an equation.

2 x^{2} + 5 x - 3 = 0

$2 x^{2} + 5 x - 3 = 0$

Step 2

Factor by grouping.

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Step 2.1

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 2 \cdot - 3 = - 6

$a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is

b = 5

$b = 5$ .

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Step 2.1.1

Factor

5

$5$ out of

5 x

$5 x$ .

2 x^{2} + 5 (x) - 3 = 0

$2 x^{2} + 5 (x) - 3 = 0$

Step 2.1.2

Rewrite

5

$5$ as

- 1

$- 1$ plus

6

$6$

2 x^{2} + (- 1 + 6) x - 3 = 0

$2 x^{2} + (- 1 + 6) x - 3 = 0$

Step 2.1.3

Apply the distributive property.

2 x^{2} - 1 x + 6 x - 3 = 0

$2 x^{2} - 1 x + 6 x - 3 = 0$

2 x^{2} - 1 x + 6 x - 3 = 0

$2 x^{2} - 1 x + 6 x - 3 = 0$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

(2 x^{2} - 1 x) + 6 x - 3 = 0

$(2 x^{2} - 1 x) + 6 x - 3 = 0$

Step 2.2.2

Factor out the greatest common factor (GCF) from each group.

x (2 x - 1) + 3 (2 x - 1) = 0

$x (2 x - 1) + 3 (2 x - 1) = 0$

x (2 x - 1) + 3 (2 x - 1) = 0

$x (2 x - 1) + 3 (2 x - 1) = 0$

Step 2.3

Factor the polynomial by factoring out the greatest common factor,

2 x - 1

$2 x - 1$ .

(2 x - 1) (x + 3) = 0

$(2 x - 1) (x + 3) = 0$

(2 x - 1) (x + 3) = 0

$(2 x - 1) (x + 3) = 0$

Step 3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

2 x - 1 = 0

$2 x - 1 = 0$

x + 3 = 0

$x + 3 = 0$

Step 4

Set

2 x - 1

$2 x - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

2 x - 1

$2 x - 1$ equal to

0

$0$ .

2 x - 1 = 0

$2 x - 1 = 0$

Step 4.2

Solve

2 x - 1 = 0

$2 x - 1 = 0$ for

x

$x$ .

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Step 4.2.1

Add

1

$1$ to both sides of the equation.

2 x = 1

$2 x = 1$

Step 4.2.2

Divide each term in

2 x = 1

$2 x = 1$ by

2

$2$ and simplify.

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Step 4.2.2.1

Divide each term in

2 x = 1

$2 x = 1$ by

2

$2$ .

\frac{2 x}{2} = \frac{1}{2}

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

\frac{2 x}{2} = \frac{1}{2}

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{1}{2}

$x = \frac{1}{2}$

x = \frac{1}{2}

$x = \frac{1}{2}$

x = \frac{1}{2}

$x = \frac{1}{2}$

x = \frac{1}{2}

$x = \frac{1}{2}$

x = \frac{1}{2}

$x = \frac{1}{2}$

x = \frac{1}{2}

$x = \frac{1}{2}$

Step 5

Set

x + 3

$x + 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

x + 3

$x + 3$ equal to

0

$0$ .

x + 3 = 0

$x + 3 = 0$

Step 5.2

Subtract

3

$3$ from both sides of the equation.

x = - 3

$x = - 3$

x = - 3

$x = - 3$

Step 6

The final solution is all the values that make

(2 x - 1) (x + 3) = 0

$(2 x - 1) (x + 3) = 0$ true.

x = \frac{1}{2}, - 3

$x = \frac{1}{2}, - 3$

Step 7

Use each root to create test intervals.

x < - 3

$x < - 3$

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$

x > \frac{1}{2}

$x > \frac{1}{2}$

Step 8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 8.1

Test a value on the interval

x < - 3

$x < - 3$ to see if it makes the inequality true.

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Step 8.1.1

Choose a value on the interval

x < - 3

$x < - 3$ and see if this value makes the original inequality true.

x = - 6

$x = - 6$

Step 8.1.2

Replace

x

$x$ with

- 6

$- 6$ in the original inequality.

2 {(- 6)}^{2} + 5 (- 6) - 3 < 0

$2 {(- 6)}^{2} + 5 (- 6) - 3 < 0$

Step 8.1.3

The left side

39

$39$ is not less than the right side

0

$0$ , which means that the given statement is false.

False

Step 8.2

Test a value on the interval

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

x = 0

$x = 0$

Step 8.2.2

Replace

x

$x$ with

0

$0$ in the original inequality.

2 {(0)}^{2} + 5 (0) - 3 < 0

$2 {(0)}^{2} + 5 (0) - 3 < 0$

Step 8.2.3

The left side

- 3

$- 3$ is less than the right side

0

$0$ , which means that the given statement is always true.

True

Step 8.3

Test a value on the interval

x > \frac{1}{2}

$x > \frac{1}{2}$ to see if it makes the inequality true.

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Step 8.3.1

Choose a value on the interval

x > \frac{1}{2}

$x > \frac{1}{2}$ and see if this value makes the original inequality true.

x = 3

$x = 3$

Step 8.3.2

Replace

x

$x$ with

3

$3$ in the original inequality.

2 {(3)}^{2} + 5 (3) - 3 < 0

$2 {(3)}^{2} + 5 (3) - 3 < 0$

Step 8.3.3

The left side

30

$30$ is not less than the right side

0

$0$ , which means that the given statement is false.

False

Step 8.4

Compare the intervals to determine which ones satisfy the original inequality.

x < - 3

$x < - 3$ False

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$ True

x > \frac{1}{2}

$x > \frac{1}{2}$ False

x < - 3

$x < - 3$ False

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$ True

x > \frac{1}{2}

$x > \frac{1}{2}$ False

Step 9

The solution consists of all of the true intervals.

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

- 3 < x < \frac{1}{2}

$- 3 < x < \frac{1}{2}$

Interval Notation:

(- 3, \frac{1}{2})

$(- 3, \frac{1}{2})$

Step 11