Solve the Inequality for x 3x^2-8x+5>=0

$3 x^{2} - 8 x + 5 \geq 0$

Step 1

Convert the inequality to an equation.

$3 x^{2} - 8 x + 5 = 0$

Step 2

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 $b = - 8$ .

Factor $- 8$ out of $- 8 x$ .

$3 x^{2} - 8 x + 5 = 0$

Rewrite $- 8$ as $- 3$ plus $- 5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Step 3

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

$x - 1 = 0$

$3 x - 5 = 0$

Step 4

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $3 x - 5$ equal to $0$ and solve for $x$ .

Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Solve $3 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$3 x = 5$

Divide each term in $3 x = 5$ by $3$ and simplify.

Divide each term in $3 x = 5$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 6

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

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

Step 7

Use each root to create test intervals.

$x < 1$

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

$x > \frac{5}{3}$

Step 8

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 $x < 1$ to see if it makes the inequality true.

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$3 {(0)}^{2} - 8 \cdot 0 + 5 \geq 0$

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 $1 < x < \frac{5}{3}$ to see if it makes the inequality true.

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

$x = 1.33$

Replace $x$ with $1.33$ in the original inequality.

$3 {(1.33)}^{2} - 8 \cdot 1.33 + 5 \geq 0$

The left side $- 0.3333$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$3 {(4)}^{2} - 8 \cdot 4 + 5 \geq 0$

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < 1$ True

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

$x > \frac{5}{3}$ True

$x < 1$ True

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

$x > \frac{5}{3}$ True

Step 9

The solution consists of all of the true intervals.

$x \leq 1$ or $x \geq \frac{5}{3}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x \leq 1 or x \geq \frac{5}{3}$

Interval Notation:

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

Step 11

Solve the Inequality for x 3x^2-8x+5>=0

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