Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary.

The solution set is
Tロ,

Ask by Kelly Reid. in the United States

Mar 19,2025

Question

Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary. $$ 4 w^{2}-9 \geq 0 $$ The solution set is $$ \square . ~ $$ Tロ, $$ \square] $$

UpStudy AI · Accepted Answer

Solve the quadratic equation by following steps: - step0: Solve using square roots: $$4w^{2}-9=0$$ - step1: Add or subtract both sides: $$4w^{2}=9$$ - step2: Divide both sides: $$\frac{4w^{2}}{4}=\frac{9}{4}$$ - step3: Divide the numbers: $$w^{2}=\frac{9}{4}$$ - step4: Simplify the expression: $$w=\pm \sqrt{\frac{9}{4}}$$ - step5: Simplify the expression: $$w=\pm \frac{3}{2}$$ - step6: Separate into possible cases: $$\begin{align}&w=\frac{3}{2}\&w=-\frac{3}{2}\end{align}$$ - step7: Rewrite: $$w_{1}=-\frac{3}{2},w_{2}=\frac{3}{2}$$ Solve the equation $$ 4w^{2}-9>0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$4w^{2}-9>0$$ - step1: Rewrite the expression: $$4w^{2}-9=0$$ - step2: Move the constant to the right side: $$4w^{2}=0+9$$ - step3: Remove 0: $$4w^{2}=9$$ - step4: Divide both sides: $$\frac{4w^{2}}{4}=\frac{9}{4}$$ - step5: Divide the numbers: $$w^{2}=\frac{9}{4}$$ - step6: Simplify the expression: $$w=\pm \sqrt{\frac{9}{4}}$$ - step7: Simplify: $$w=\pm \frac{3}{2}$$ - step8: Separate into possible cases: $$\begin{align}&w=\frac{3}{2}\&w=-\frac{3}{2}\end{align}$$ - step9: Determine the test intervals: $$\begin{align}&w<-\frac{3}{2}\&-\frac{3}{2}\frac{3}{2}\end{align}$$ - step10: Choose a value: $$\begin{align}&w_{1}=-3\&w_{2}=0\&w_{3}=3\end{align}$$ - step11: Test the chosen value: $$\begin{align}&w<-\frac{3}{2}\textrm{ }\textrm{is the solution}\&-\frac{3}{2}\frac{3}{2}\textrm{ }\textrm{is the solution}\end{align}$$ - step12: The final solution is $$w \in \left(-\infty,-\frac{3}{2}\right)\cup \left(\frac{3}{2},+\infty\right):$$ $$w \in \left(-\infty,-\frac{3}{2}\right)\cup \left(\frac{3}{2},+\infty\right)$$ To solve the inequality $$ 4w^{2} - 9 \geq 0 $$, we first find the points where the expression equals zero: 1. Set the equation to zero: $$ 4w^{2} - 9 = 0 $$ 2. Solve for $$ w $$: $$ 4w^{2} = 9 \implies w^{2} = \frac{9}{4} \implies w = \pm \frac{3}{2} $$ Thus, the critical points are $$ w = -\frac{3}{2} $$ and $$ w = \frac{3}{2} $$. Next, we analyze the intervals determined by these critical points to find where the inequality $$ 4w^{2} - 9 \geq 0 $$ holds true. 3. The intervals to test are: - $$ (-\infty, -\frac{3}{2}) $$ - $$ (-\frac{3}{2}, \frac{3}{2}) $$ - $$ (\frac{3}{2}, +\infty) $$ 4. Choose test points from each interval: - For $$ w < -\frac{3}{2} $$ (e.g., $$ w = -2 $$): $$ 4(-2)^{2} - 9 = 16 - 9 = 7 \quad (\text{True}) $$ - For $$ -\frac{3}{2} < w < \frac{3}{2} $$ (e.g., $$ w = 0 $$): $$ 4(0)^{2} - 9 = -9 \quad (\text{False}) $$ - For $$ w > \frac{3}{2} $$ (e.g., $$ w = 2 $$): $$ 4(2)^{2} - 9 = 16 - 9 = 7 \quad (\text{True}) $$ 5. The inequality $$ 4w^{2} - 9 \geq 0 $$ holds in the intervals $$ (-\infty, -\frac{3}{2}] $$ and $$ [\frac{3}{2}, +\infty) $$. Thus, the solution set in interval notation is: $$ (-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, +\infty) $$ The final answer is: $$ \text{The solution set is } (-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, +\infty) $$ The solution set is $$ (-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, +\infty) $$.

Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary.

The solution set is
Tロ,

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Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary. The solution set is Tロ,