Convert to Interval Notation 4x^2+5>x^2+16x

$4 x^{2} + 5 > x^{2} + 16 x$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$x^{2} + 16 x < 4 x^{2} + 5$

Step 2

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

Subtract $4 x^{2}$ from both sides of the inequality.

$x^{2} + 16 x - 4 x^{2} < 5$

Subtract $4 x^{2}$ from $x^{2}$ .

$- 3 x^{2} + 16 x < 5$

Step 3

Convert the inequality to an equation.

$- 3 x^{2} + 16 x = 5$

Step 4

Subtract $5$ from both sides of the equation.

$- 3 x^{2} + 16 x - 5 = 0$

Step 5

Factor the left side of the equation.

Factor $- 1$ out of $- 3 x^{2} + 16 x - 5$ .

Factor $- 1$ out of $- 3 x^{2}$ .

$- (3 x^{2}) + 16 x - 5 = 0$

Factor $- 1$ out of $16 x$ .

$- (3 x^{2}) - (- 16 x) - 5 = 0$

Rewrite $- 5$ as $- 1 (5)$ .

$- (3 x^{2}) - (- 16 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (3 x^{2}) - (- 16 x)$ .

$- (3 x^{2} - 16 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (3 x^{2} - 16 x) - 1 (5)$ .

$- (3 x^{2} - 16 x + 5) = 0$

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 $a \cdot c = 3 \cdot 5 = 15$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 x$ .

$- (3 x^{2} - 16 x + 5) = 0$

Rewrite $- 16$ as $- 1$ plus $- 15$

$- (3 x^{2} + (- 1 - 15) x + 5) = 0$

Apply the distributive property.

$- (3 x^{2} - 1 x - 15 x + 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((3 x^{2} - 1 x) - 15 x + 5) = 0$

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

$- (x (3 x - 1) - 5 (3 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$- ((3 x - 1) (x - 5)) = 0$

Remove unnecessary parentheses.

$- (3 x - 1) (x - 5) = 0$

Step 6

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

$3 x - 1 = 0$

$x - 5 = 0$

Step 7

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 8

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 9

The final solution is all the values that make $- (3 x - 1) (x - 5) = 0$ true.

$x = \frac{1}{3}, 5$

Step 10

Use each root to create test intervals.

$x < \frac{1}{3}$

$\frac{1}{3} < x < 5$

$x > 5$

Step 11

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 $x < \frac{1}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$4 {(0)}^{2} + 5 > {(0)}^{2} + 16 (0)$

The left side $5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{1}{3} < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{3} < x < 5$ and see if this value makes the original inequality true.

$x = 3$

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

$4 {(3)}^{2} + 5 > {(3)}^{2} + 16 (3)$

The left side $41$ is not greater than the right side $57$ , 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.

$4 {(8)}^{2} + 5 > {(8)}^{2} + 16 (8)$

The left side $261$ 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.

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 5$ False

$x > 5$ True

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 5$ False

$x > 5$ True

Step 12

The solution consists of all of the true intervals.

$x < \frac{1}{3}$ or $x > 5$

Step 13

Convert the inequality to interval notation.

$(- \infty, \frac{1}{3}) \cup (5, \infty)$

Step 14

Convert to Interval Notation 4x^2+5>x^2+16x

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