Solve -6x^2≤-16

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, - \frac{2 6}{3}] \cup [\frac{2 6}{3}, + \infty)

Evaluate

- 6 x^{2} \leq - 16

Move the expression to the left side

- 6 x^{2} - (- 16) \leq 0

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

- 6 x^{2} + 16 \leq 0

Rewrite the expression

- 6 x^{2} + 16 = 0

Move the constant to the right-hand side and change its sign

- 6 x^{2} = 0 - 16

Removing 0 doesn't change the value,so remove it from the expression

- 6 x^{2} = - 16

Change the signs on both sides of the equation

6 x^{2} = 16

Divide both sides

\frac{6 x ^{2}}{6} = \frac{16}{6}

Divide the numbers

x^{2} = \frac{16}{6}

Cancel out the common factor 2

x^{2} = \frac{8}{3}

Take the root of both sides of the equation and remember to use both positive and negative roots

x = \pm \frac{8}{3}

Simplify the expression

More Steps

x = \pm \frac{2 6}{3}

Separate the equation into 2 possible cases

x = \frac{2 6}{3} x = - \frac{2 6}{3}

Determine the test intervals using the critical values

x < - \frac{2 6}{3} - \frac{2 6}{3} < x < \frac{2 6}{3} x > \frac{2 6}{3}

Choose a value form each interval

x_{1} = - 3 x_{2} = 0 x_{3} = 3

To determine if x < - \frac{2 6}{3} is the solution to the inequality,test if the chosen value x = - 3 satisfies the initial inequality

More Steps

x < - \frac{2 6}{3} is the solution x_{2} = 0 x_{3} = 3

To determine if - \frac{2 6}{3} < x < \frac{2 6}{3} is the solution to the inequality,test if the chosen value x = 0 satisfies the initial inequality

More Steps

x < - \frac{2 6}{3} is the solution - \frac{2 6}{3} < x < \frac{2 6}{3} is not a solution x_{3} = 3

To determine if x > \frac{2 6}{3} is the solution to the inequality,test if the chosen value x = 3 satisfies the initial inequality

More Steps

x < - \frac{2 6}{3} is the solution - \frac{2 6}{3} < x < \frac{2 6}{3} is not a solution x > \frac{2 6}{3} is the solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x \leq - \frac{2 6}{3} is the solution x \geq \frac{2 6}{3} is the solution

Solution

x \in (- \infty, - \frac{2 6}{3}] \cup [\frac{2 6}{3}, + \infty)

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