Solve 20≤g^219 - ScanMath

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for g

g \in (- \infty, - \frac{2 95}{19}] \cup [\frac{2 95}{19}, + \infty)

Evaluate

20 \leq g^{2} \times 19

Use the commutative property to reorder the terms

20 \leq 19 g^{2}

Move the expression to the left side

20 - 19 g^{2} \leq 0

Rewrite the expression

20 - 19 g^{2} = 0

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

- 19 g^{2} = 0 - 20

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

- 19 g^{2} = - 20

Change the signs on both sides of the equation

19 g^{2} = 20

Divide both sides

\frac{19 g ^{2}}{19} = \frac{20}{19}

Divide the numbers

g^{2} = \frac{20}{19}

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

g = \pm \frac{20}{19}

Simplify the expression

More Steps

g = \pm \frac{2 95}{19}

Separate the equation into 2 possible cases

g = \frac{2 95}{19} g = - \frac{2 95}{19}

Determine the test intervals using the critical values

g < - \frac{2 95}{19} - \frac{2 95}{19} < g < \frac{2 95}{19} g > \frac{2 95}{19}

Choose a value form each interval

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

To determine if g < - \frac{2 95}{19} is the solution to the inequality,test if the chosen value g = - 2 satisfies the initial inequality

More Steps

g < - \frac{2 95}{19} is the solution g_{2} = 0 g_{3} = 2

To determine if - \frac{2 95}{19} < g < \frac{2 95}{19} is the solution to the inequality,test if the chosen value g = 0 satisfies the initial inequality

More Steps

g < - \frac{2 95}{19} is the solution - \frac{2 95}{19} < g < \frac{2 95}{19} is not a solution g_{3} = 2

To determine if g > \frac{2 95}{19} is the solution to the inequality,test if the chosen value g = 2 satisfies the initial inequality

More Steps

g < - \frac{2 95}{19} is the solution - \frac{2 95}{19} < g < \frac{2 95}{19} is not a solution g > \frac{2 95}{19} is the solution

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

g \leq - \frac{2 95}{19} is the solution g \geq \frac{2 95}{19} is the solution

Solution

g \in (- \infty, - \frac{2 95}{19}] \cup [\frac{2 95}{19}, + \infty)

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