Solve the Inequality for x 5x<=12-2x^2

$5 x \leq 12 - 2 x^{2}$

Add $2 x^{2}$ to both sides of the inequality.

$5 x + 2 x^{2} \leq 12$

Convert the inequality to an equation.

$5 x + 2 x^{2} = 12$

Subtract $12$ from both sides of the equation.

$5 x + 2 x^{2} - 12 = 0$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$5 u + 2 u^{2} - 12 = 0$

Factor by grouping.

Reorder terms.

$2 u^{2} + 5 u - 12 = 0$

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 $b = 5$ .

Factor $5$ out of $5 u$ .

$2 u^{2} + 5 (u) - 12 = 0$

Rewrite $5$ as $- 3$ plus $8$

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

Apply the distributive property.

$2 u^{2} - 3 u + 8 u - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$u (2 u - 3) + 4 (2 u - 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 3$ .

$(2 u - 3) (u + 4) = 0$

Replace all occurrences of $u$ with $x$ .

$(2 x - 3) (x + 4) = 0$

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

$2 x - 3 = 0$

$x + 4 = 0$

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

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

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

Use each root to create test intervals.

$x < - 4$

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

$x > \frac{3}{2}$

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 $x < - 4$ 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.

$5 (- 6) \leq 12 - 2 {(- 6)}^{2}$

The left side $- 30$ is greater than the right side $- 60$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$5 (0) \leq 12 - 2 {(0)}^{2}$

The left side $0$ is less than the right side $12$ , which means that the given statement is always true.

True

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

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

$x = 4$

Replace $x$ with $4$ in the original inequality.

$5 (4) \leq 12 - 2 {(4)}^{2}$

The left side $20$ is greater than the right side $- 20$ , which means that the given statement is false.

False

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

$x < - 4$ False

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

$x > \frac{3}{2}$ False

$x < - 4$ False

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

$x > \frac{3}{2}$ False

The solution consists of all of the true intervals.

$- 4 \leq x \leq \frac{3}{2}$

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x \leq \frac{3}{2}$

Interval Notation:

$[- 4, \frac{3}{2}]$

Solve the Inequality for x 5x<=12-2x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.