Convert to Interval Notation 42x^2-13x>=42

$42 x^{2} - 13 x \geq 42$

Step 1

Convert the inequality to an equation.

$42 x^{2} - 13 x = 42$

Step 2

Subtract $42$ from both sides of the equation.

$42 x^{2} - 13 x - 42 = 0$

Step 3

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 $a \cdot c = 42 \cdot - 42 = - 1764$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$42 x^{2} - 13 x - 42 = 0$

Rewrite $- 13$ as $36$ plus $- 49$

$42 x^{2} + (36 - 49) x - 42 = 0$

Apply the distributive property.

$42 x^{2} + 36 x - 49 x - 42 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(42 x^{2} + 36 x) - 49 x - 42 = 0$

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

$6 x (7 x + 6) - 7 (7 x + 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x + 6$ .

$(7 x + 6) (6 x - 7) = 0$

Step 4

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

$7 x + 6 = 0$

$6 x - 7 = 0$

Step 5

Set $7 x + 6$ equal to $0$ and solve for $x$ .

Set $7 x + 6$ equal to $0$ .

$7 x + 6 = 0$

Solve $7 x + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$7 x = - 6$

Divide each term in $7 x = - 6$ by $7$ and simplify.

Divide each term in $7 x = - 6$ by $7$ .

$\frac{7 x}{7} = \frac{- 6}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 6}{7}$

Divide $x$ by $1$ .

$x = \frac{- 6}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{6}{7}$

Step 6

Set $6 x - 7$ equal to $0$ and solve for $x$ .

Set $6 x - 7$ equal to $0$ .

$6 x - 7 = 0$

Solve $6 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$6 x = 7$

Divide each term in $6 x = 7$ by $6$ and simplify.

Divide each term in $6 x = 7$ by $6$ .

$\frac{6 x}{6} = \frac{7}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{7}{6}$

Divide $x$ by $1$ .

$x = \frac{7}{6}$

Step 7

The final solution is all the values that make $(7 x + 6) (6 x - 7) = 0$ true.

$x = - \frac{6}{7}, \frac{7}{6}$

Step 8

Use each root to create test intervals.

$x < - \frac{6}{7}$

$- \frac{6}{7} < x < \frac{7}{6}$

$x > \frac{7}{6}$

Step 9

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 $x < - \frac{6}{7}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{6}{7}$ and see if this value makes the original inequality true.

$x = - 3$

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

$42 {(- 3)}^{2} - 13 \cdot - 3 \geq 42$

The left side $417$ is greater than the right side $42$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{6}{7} < x < \frac{7}{6}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{6}{7} < x < \frac{7}{6}$ and see if this value makes the original inequality true.

$x = 0$

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

$42 {(0)}^{2} - 13 \cdot 0 \geq 42$

The left side $0$ is less than the right side $42$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{7}{6}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{7}{6}$ and see if this value makes the original inequality true.

$x = 4$

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

$42 {(4)}^{2} - 13 \cdot 4 \geq 42$

The left side $620$ is greater than the right side $42$ , which means that the given statement is always true.

True

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

$x < - \frac{6}{7}$ True

$- \frac{6}{7} < x < \frac{7}{6}$ False

$x > \frac{7}{6}$ True

$x < - \frac{6}{7}$ True

$- \frac{6}{7} < x < \frac{7}{6}$ False

$x > \frac{7}{6}$ True

Step 10

The solution consists of all of the true intervals.

$x \leq - \frac{6}{7}$ or $x \geq \frac{7}{6}$

Step 11

Convert the inequality to interval notation.

$(- \infty, - \frac{6}{7}] \cup [\frac{7}{6}, \infty)$

Step 12

Convert to Interval Notation 42x^2-13x>=42

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