Solve for x 5(x^2-1)>24x

$5 (x^{2} - 1) > 24 x$

Step 1

Simplify $5 (x^{2} - 1)$ .

Rewrite.

$0 + 0 + 5 (x^{2} - 1) > 24 x$

Simplify by adding zeros.

$5 (x^{2} - 1) > 24 x$

Apply the distributive property.

$5 x^{2} + 5 \cdot - 1 > 24 x$

Multiply $5$ by $- 1$ .

$5 x^{2} - 5 > 24 x$

Step 2

Subtract $24 x$ from both sides of the inequality.

$5 x^{2} - 5 - 24 x > 0$

Step 3

Convert the inequality to an equation.

$5 x^{2} - 5 - 24 x = 0$

Step 4

Factor by grouping.

Reorder terms.

$5 x^{2} - 24 x - 5 = 0$

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 $a \cdot c = 5 \cdot - 5 = - 25$ and whose sum is $b = - 24$ .

Factor $- 24$ out of $- 24 x$ .

$5 x^{2} - 24 x - 5 = 0$

Rewrite $- 24$ as $1$ plus $- 25$

$5 x^{2} + (1 - 25) x - 5 = 0$

Apply the distributive property.

$5 x^{2} + 1 x - 25 x - 5 = 0$

Multiply $x$ by $1$ .

$5 x^{2} + x - 25 x - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{2} + x) - 25 x - 5 = 0$

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

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

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

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

Step 5

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

$5 x + 1 = 0$

$x - 5 = 0$

Step 6

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.

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

Step 7

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.

$x = 5$

Step 8

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

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

Step 9

Use each root to create test intervals.

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

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

$x > 5$

Step 10

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.

$5 ({(- 3)}^{2} - 1) > 24 (- 3)$

The left side $40$ is greater than the right side $- 72$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$5 ({(0)}^{2} - 1) > 24 (0)$

The left side $- 5$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

$5 ({(8)}^{2} - 1) > 24 (8)$

The left side $315$ is greater than the right side $192$ , 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

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

$x > 5$ True

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

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

$x > 5$ True

Step 11

The solution consists of all of the true intervals.

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

Step 12

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 13

Solve for x 5(x^2-1)>24x

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