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Answer
The solution is \( x \leq -4 \) or \( x \geq \frac{5}{2} \).
Solution
Solve the equation \( 2x^{2}+3x-20 \geq 0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(2x^{2}+3x-20\geq 0\)
- step1: Rewrite the expression:
\(2x^{2}+3x-20=0\)
- step2: Factor the expression:
\(\left(x+4\right)\left(2x-5\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&x+4=0\\&2x-5=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=-4\\&x=\frac{5}{2}\end{align}\)
- step5: Determine the test intervals:
\(\begin{align}&x<-4\\&-4\frac{5}{2}\end{align}\)
- step6: Choose a value:
\(\begin{align}&x_{1}=-5\\&x_{2}=-1\\&x_{3}=4\end{align}\)
- step7: Test the chosen value:
\(\begin{align}&x<-4\textrm{ }\textrm{is the solution}\\&-4\frac{5}{2}\textrm{ }\textrm{is the solution}\end{align}\)
- step8: Include the critical value:
\(\begin{align}&x\leq -4\textrm{ }\textrm{is the solution}\\&x\geq \frac{5}{2}\textrm{ }\textrm{is the solution}\end{align}\)
- step9: The final solution is \(x \in \left(-\infty,-4\right]\cup \left[\frac{5}{2},+\infty\right):\)
\(x \in \left(-\infty,-4\right]\cup \left[\frac{5}{2},+\infty\right)\)
The solution to the inequality \(2x^{2}+3x-20 \geq 0\) is \(x \in (-\infty, -4) \cup \left[\frac{5}{2}, +\infty)\).
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Simplify this solution Extra Insights
To solve the inequality \( 2x^{2}+3x-20 \geq 0 \), first, we find the roots by using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 3 \), and \( c = -20 \). Plugging those values in gives you the roots which can be used to determine the intervals on a number line. Once you have the roots, test the intervals defined by them. These intervals will tell you where the inequality holds true. Keep an eye on including or excluding the roots based on whether or not the expression equals zero in the inequality. You’ll find that understanding the behavior of quadratics helps visualize the solution!
