Solve the Inequality for x -2x^2+13x-20<0

$- 2 x^{2} + 13 x - 20 < 0$

Step 1

Convert the inequality to an equation.

$- 2 x^{2} + 13 x - 20 = 0$

Step 2

Factor the left side of the equation.

Factor $- 1$ out of $- 2 x^{2} + 13 x - 20$ .

Factor $- 1$ out of $- 2 x^{2}$ .

$- (2 x^{2}) + 13 x - 20 = 0$

Factor $- 1$ out of $13 x$ .

$- (2 x^{2}) - (- 13 x) - 20 = 0$

Rewrite $- 20$ as $- 1 (20)$ .

$- (2 x^{2}) - (- 13 x) - 1 \cdot 20 = 0$

Factor $- 1$ out of $- (2 x^{2}) - (- 13 x)$ .

$- (2 x^{2} - 13 x) - 1 \cdot 20 = 0$

Factor $- 1$ out of $- (2 x^{2} - 13 x) - 1 (20)$ .

$- (2 x^{2} - 13 x + 20) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 20 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$- (2 x^{2} - 13 x + 20) = 0$

Rewrite $- 13$ as $- 5$ plus $- 8$

$- (2 x^{2} + (- 5 - 8) x + 20) = 0$

Apply the distributive property.

$- (2 x^{2} - 5 x - 8 x + 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((2 x^{2} - 5 x) - 8 x + 20) = 0$

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

$- (x (2 x - 5) - 4 (2 x - 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$- ((2 x - 5) (x - 4)) = 0$

Remove unnecessary parentheses.

$- (2 x - 5) (x - 4) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 5 = 0$

$x - 4 = 0$

Step 4

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

The final solution is all the values that make $- (2 x - 5) (x - 4) = 0$ true.

$x = \frac{5}{2}, 4$

Step 7

Use each root to create test intervals.

$x < \frac{5}{2}$

$\frac{5}{2} < x < 4$

$x > 4$

Step 8

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

Test a value on the interval $x < \frac{5}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{5}{2}$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$- 2 {(0)}^{2} + 13 (0) - 20 < 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{5}{2} < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $\frac{5}{2} < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$- 2 {(3)}^{2} + 13 (3) - 20 < 0$

The left side $1$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$- 2 {(6)}^{2} + 13 (6) - 20 < 0$

The left side $- 14$ is less than the right side $0$ , which means that the given statement is always true.

True

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

$x < \frac{5}{2}$ True

$\frac{5}{2} < x < 4$ False

$x > 4$ True

$x < \frac{5}{2}$ True

$\frac{5}{2} < x < 4$ False

$x > 4$ True

Step 9

The solution consists of all of the true intervals.

$x < \frac{5}{2}$ or $x > 4$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{5}{2} or x > 4$

Interval Notation:

$(- \infty, \frac{5}{2}) \cup (4, \infty)$

Step 11

Solve the Inequality for x -2x^2+13x-20<0

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