Solve the Inequality for t 2t^2+11t+12<0

$2 t^{2} + 11 t + 12 < 0$

Convert the inequality to an equation.

$2 t^{2} + 11 t + 12 = 0$

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 $a \cdot c = 2 \cdot 12 = 24$ and whose sum is $b = 11$ .

Factor $11$ out of $11 t$ .

$2 t^{2} + 11 (t) + 12 = 0$

Rewrite $11$ as $3$ plus $8$

$2 t^{2} + (3 + 8) t + 12 = 0$

Apply the distributive property.

$2 t^{2} + 3 t + 8 t + 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 t^{2} + 3 t) + 8 t + 12 = 0$

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

$t (2 t + 3) + 4 (2 t + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 t + 3$ .

$(2 t + 3) (t + 4) = 0$

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

$2 t + 3 = 0$

$t + 4 = 0$

Set $2 t + 3$ equal to $0$ and solve for $t$ .

Set $2 t + 3$ equal to $0$ .

$2 t + 3 = 0$

Solve $2 t + 3 = 0$ for $t$ .

Subtract $3$ from both sides of the equation.

$2 t = - 3$

Divide each term in $2 t = - 3$ by $2$ and simplify.

Divide each term in $2 t = - 3$ by $2$ .

$\frac{2 t}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 t}{2} = \frac{- 3}{2}$

Divide $t$ by $1$ .

$t = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{3}{2}$

Set $t + 4$ equal to $0$ and solve for $t$ .

Set $t + 4$ equal to $0$ .

$t + 4 = 0$

Subtract $4$ from both sides of the equation.

$t = - 4$

The final solution is all the values that make $(2 t + 3) (t + 4) = 0$ true.

$t = - \frac{3}{2}, - 4$

Use each root to create test intervals.

$t < - 4$

$- 4 < t < - \frac{3}{2}$

$t > - \frac{3}{2}$

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 $t < - 4$ to see if it makes the inequality true.

Choose a value on the interval $t < - 4$ and see if this value makes the original inequality true.

$t = - 6$

Replace $t$ with $- 6$ in the original inequality.

$2 {(- 6)}^{2} + 11 (- 6) + 12 < 0$

The left side $18$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < t < - \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < t < - \frac{3}{2}$ and see if this value makes the original inequality true.

$t = - 3$

Replace $t$ with $- 3$ in the original inequality.

$2 {(- 3)}^{2} + 11 (- 3) + 12 < 0$

The left side $- 3$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $t > - \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $t > - \frac{3}{2}$ and see if this value makes the original inequality true.

$t = 0$

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

$2 {(0)}^{2} + 11 (0) + 12 < 0$

The left side $12$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$t < - 4$ False

$- 4 < t < - \frac{3}{2}$ True

$t > - \frac{3}{2}$ False

$t < - 4$ False

$- 4 < t < - \frac{3}{2}$ True

$t > - \frac{3}{2}$ False

The solution consists of all of the true intervals.

$- 4 < t < - \frac{3}{2}$

The result can be shown in multiple forms.

Inequality Form:

$- 4 < t < - \frac{3}{2}$

Interval Notation:

$(- 4, - \frac{3}{2})$

Solve the Inequality for t 2t^2+11t+12<0

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