Solve the Inequality for x 49x^2-42x>-9

$49 x^{2} - 42 x > - 9$

Step 1

Convert the inequality to an equation.

$49 x^{2} - 42 x = - 9$

Step 2

Add $9$ to both sides of the equation.

$49 x^{2} - 42 x + 9 = 0$

Step 3

Factor using the perfect square rule.

Rewrite $49 x^{2}$ as ${(7 x)}^{2}$ .

${(7 x)}^{2} - 42 x + 9 = 0$

Rewrite $9$ as $3^{2}$ .

${(7 x)}^{2} - 42 x + 3^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$42 x = 2 \cdot (7 x) \cdot 3$

Rewrite the polynomial.

${(7 x)}^{2} - 2 \cdot (7 x) \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 7 x$ and $b = 3$ .

${(7 x - 3)}^{2} = 0$

Step 4

Set the $7 x - 3$ equal to $0$ .

$7 x - 3 = 0$

Step 5

Solve for $x$ .

Add $3$ to both sides of the equation.

$7 x = 3$

Divide each term in $7 x = 3$ by $7$ and simplify.

Divide each term in $7 x = 3$ by $7$ .

$\frac{7 x}{7} = \frac{3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{3}{7}$

Divide $x$ by $1$ .

$x = \frac{3}{7}$

Step 6

Use each root to create test intervals.

$x < \frac{3}{7}$

$x > \frac{3}{7}$

Step 7

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 $x < \frac{3}{7}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{3}{7}$ and see if this value makes the original inequality true.

$x = 0$

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

$49 {(0)}^{2} - 42 \cdot 0 > - 9$

The left side $0$ is greater than the right side $- 9$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{3}{7}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{3}{7}$ and see if this value makes the original inequality true.

$x = 3$

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

$49 {(3)}^{2} - 42 \cdot 3 > - 9$

The left side $315$ is greater than the right side $- 9$ , which means that the given statement is always true.

True

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

$x < \frac{3}{7}$ True

$x > \frac{3}{7}$ True

$x < \frac{3}{7}$ True

$x > \frac{3}{7}$ True

Step 8

The solution consists of all of the true intervals.

$x < \frac{3}{7}$ or $x > \frac{3}{7}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{3}{7} or x > \frac{3}{7}$

Interval Notation:

$(- \infty, \frac{3}{7}) \cup (\frac{3}{7}, \infty)$

Step 10

Solve the Inequality for x 49x^2-42x>-9

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