Solve the Inequality for x 2x^2-7x+3>0

$2 x^{2} - 7 x + 3 > 0$

Step 1

Convert the inequality to an equation.

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

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

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

Rewrite $- 7$ as $- 1$ plus $- 6$

$2 x^{2} + (- 1 - 6) x + 3 = 0$

Apply the distributive property.

$2 x^{2} - 1 x - 6 x + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) - 6 x + 3 = 0$

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

$x (2 x - 1) - 3 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x - 3) = 0$

Step 3

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

$2 x - 1 = 0$

$x - 3 = 0$

Step 4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 5

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

The final solution is all the values that make $(2 x - 1) (x - 3) = 0$ true.

$x = \frac{1}{2}, 3$

Step 7

Use each root to create test intervals.

$x < \frac{1}{2}$

$\frac{1}{2} < x < 3$

$x > 3$

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 $x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{1}{2} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

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

The left side $- 3$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$2 {(6)}^{2} - 7 \cdot 6 + 3 > 0$

The left side $33$ 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.

$x < \frac{1}{2}$ True

$\frac{1}{2} < x < 3$ False

$x > 3$ True

$x < \frac{1}{2}$ True

$\frac{1}{2} < x < 3$ False

$x > 3$ True

Step 9

The solution consists of all of the true intervals.

$x < \frac{1}{2}$ or $x > 3$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{1}{2} or x > 3$

Interval Notation:

$(- \infty, \frac{1}{2}) \cup (3, \infty)$

Step 11

Solve the Inequality for x 2x^2-7x+3>0

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