Solve the Inequality for x x^2-3x-35<=-5x

$x^{2} - 3 x - 35 \leq - 5 x$

Move all terms containing $x$ to the left side of the inequality.

Add $5 x$ to both sides of the inequality.

$x^{2} - 3 x - 35 + 5 x \leq 0$

Add $- 3 x$ and $5 x$ .

$x^{2} + 2 x - 35 \leq 0$

Convert the inequality to an equation.

$x^{2} + 2 x - 35 = 0$

Factor $x^{2} + 2 x - 35$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$(x - 5) (x + 7) = 0$

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

$x - 5 = 0$

$x + 7 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 5$

$x > 5$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

${(- 10)}^{2} - 3 \cdot - 10 - 35 \leq - 5 \cdot - 10$

The left side $95$ is greater than the right side $50$ , which means that the given statement is false.

False

Test a value on the interval $- 7 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 3 \cdot 0 - 35 \leq - 5 \cdot 0$

The left side $- 35$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 3 \cdot 8 - 35 \leq - 5 \cdot 8$

The left side $5$ is greater than the right side $- 40$ , which means that the given statement is false.

False

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

$x < - 7$ False

$- 7 < x < 5$ True

$x > 5$ False

$x < - 7$ False

$- 7 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$- 7 \leq x \leq 5$

The result can be shown in multiple forms.

Inequality Form:

$- 7 \leq x \leq 5$

Interval Notation:

$[- 7, 5]$

Solve the Inequality for x x^2-3x-35<=-5x

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