Solve the Inequality for x -5x^2-1<=-9x-3

$- 5 x^{2} - 1 \leq - 9 x - 3$

Add $9 x$ to both sides of the inequality.

$- 5 x^{2} - 1 + 9 x \leq - 3$

Convert the inequality to an equation.

$- 5 x^{2} - 1 + 9 x = - 3$

Add $3$ to both sides of the equation.

$- 5 x^{2} - 1 + 9 x + 3 = 0$

Add $- 1$ and $3$ .

$- 5 x^{2} + 9 x + 2 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 5 x^{2} + 9 x + 2$ .

Factor $- 1$ out of $- 5 x^{2}$ .

$- (5 x^{2}) + 9 x + 2 = 0$

Factor $- 1$ out of $9 x$ .

$- (5 x^{2}) - (- 9 x) + 2 = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$- (5 x^{2}) - (- 9 x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (5 x^{2}) - (- 9 x)$ .

$- (5 x^{2} - 9 x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (5 x^{2} - 9 x) - 1 (- 2)$ .

$- (5 x^{2} - 9 x - 2) = 0$

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 $a \cdot c = 5 \cdot - 2 = - 10$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$- (5 x^{2} - 9 x - 2) = 0$

Rewrite $- 9$ as $1$ plus $- 10$

$- (5 x^{2} + (1 - 10) x - 2) = 0$

Apply the distributive property.

$- (5 x^{2} + 1 x - 10 x - 2) = 0$

Multiply $x$ by $1$ .

$- (5 x^{2} + x - 10 x - 2) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((5 x^{2} + x) - 10 x - 2) = 0$

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

$- (x (5 x + 1) - 2 (5 x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

$- ((5 x + 1) (x - 2)) = 0$

Remove unnecessary parentheses.

$- (5 x + 1) (x - 2) = 0$

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

$5 x + 1 = 0$

$x - 2 = 0$

Set $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $- (5 x + 1) (x - 2) = 0$ true.

$x = - \frac{1}{5}, 2$

Use each root to create test intervals.

$x < - \frac{1}{5}$

$- \frac{1}{5} < x < 2$

$x > 2$

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 $x < - \frac{1}{5}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{1}{5}$ and see if this value makes the original inequality true.

$x = - 3$

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

$- 5 {(- 3)}^{2} - 1 \leq - 9 \cdot - 3 - 3$

The left side $- 46$ is less than the right side $24$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1}{5} < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{5} < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$- 5 {(0)}^{2} - 1 \leq - 9 \cdot 0 - 3$

The left side $- 1$ is greater than the right side $- 3$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$- 5 {(4)}^{2} - 1 \leq - 9 \cdot 4 - 3$

The left side $- 81$ is less than the right side $- 39$ , which means that the given statement is always true.

True

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

$x < - \frac{1}{5}$ True

$- \frac{1}{5} < x < 2$ False

$x > 2$ True

$x < - \frac{1}{5}$ True

$- \frac{1}{5} < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq - \frac{1}{5}$ or $x \geq 2$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{1}{5} or x \geq 2$

Interval Notation:

$(- \infty, - \frac{1}{5}] \cup [2, \infty)$

Solve the Inequality for x -5x^2-1<=-9x-3

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