Solve the Inequality for x -2x^2-8x+7<=-3x+4

$- 2 x^{2} - 8 x + 7 \leq - 3 x + 4$

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

Add $3 x$ to both sides of the inequality.

$- 2 x^{2} - 8 x + 7 + 3 x \leq 4$

Add $- 8 x$ and $3 x$ .

$- 2 x^{2} - 5 x + 7 \leq 4$

Convert the inequality to an equation.

$- 2 x^{2} - 5 x + 7 = 4$

Subtract $4$ from both sides of the equation.

$- 2 x^{2} - 5 x + 7 - 4 = 0$

Subtract $4$ from $7$ .

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

Factor the left side of the equation.

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

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

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

Factor $- 1$ out of $- 5 x$ .

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

Rewrite $3$ as $- 1 (- 3)$ .

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

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

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

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

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

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ 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, $2 x - 1$ .

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

Remove unnecessary parentheses.

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

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

$2 x - 1 = 0$

$x + 3 = 0$

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 $1$ .

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

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

Use each root to create test intervals.

$x < - 3$

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

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

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 $x < - 3$ 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.

$- 2 {(- 6)}^{2} - 8 \cdot - 6 + 7 \leq - 3 \cdot - 6 + 4$

The left side $- 17$ is less than the right side $22$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$- 2 {(0)}^{2} - 8 \cdot 0 + 7 \leq - 3 \cdot 0 + 4$

The left side $7$ is greater than the right side $4$ , which means that the given statement is false.

False

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.

$x = 3$

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

$- 2 {(3)}^{2} - 8 \cdot 3 + 7 \leq - 3 \cdot 3 + 4$

The left side $- 35$ is less than the right side $- 5$ , which means that the given statement is always true.

True

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

$x < - 3$ True

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

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

$x < - 3$ True

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

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

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq \frac{1}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x \geq \frac{1}{2}$

Interval Notation:

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

Solve the Inequality for x -2x^2-8x+7<=-3x+4

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