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a. \( x^{2}+3x+2<0 \): \(-2 < x < -1\)
b. \( 3x^{2}-9x+6>0 \): \( x < 1 \) o \( x > 2 \)
c. \( 5x^{2}-8x+3<0 \): \( \frac{3}{5} < x < 1 \)
d. \( 8x^{2}+14x<-6 \): \(-1 < x < -\frac{3}{4}\)
e. \( 2x^{2}+3x-4 \leq 0 \): \( -\frac{\sqrt{41}+3}{4} \leq x \leq \frac{\sqrt{41}-3}{4} \)
f. \( 4x^{2}+7x-3 \geq 0 \): \( x \leq -\frac{\sqrt{97}+7}{8} \) o \( x \geq \frac{\sqrt{97}-7}{8} \)
Solución
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(8x^{2}+14x<-6\)
- step1: Move the expression to the left side:
\(8x^{2}+14x-\left(-6\right)<0\)
- step2: Remove the parentheses:
\(8x^{2}+14x+6<0\)
- step3: Rewrite the expression:
\(8x^{2}+14x+6=0\)
- step4: Factor the expression:
\(2\left(x+1\right)\left(4x+3\right)=0\)
- step5: Divide the terms:
\(\left(x+1\right)\left(4x+3\right)=0\)
- step6: Separate into possible cases:
\(\begin{align}&x+1=0\\&4x+3=0\end{align}\)
- step7: Solve the equation:
\(\begin{align}&x=-1\\&x=-\frac{3}{4}\end{align}\)
- step8: Determine the test intervals:
\(\begin{align}&x<-1\\&-1-\frac{3}{4}\end{align}\)
- step9: Choose a value:
\(\begin{align}&x_{1}=-2\\&x_{2}=-\frac{7}{8}\\&x_{3}=0\end{align}\)
- step10: Test the chosen value:
\(\begin{align}&x<-1\textrm{ }\textrm{is not a solution}\\&-1-\frac{3}{4}\textrm{ }\textrm{is not a solution}\end{align}\)
- step11: The final solution is \(-11\end{align}\)
- step6: Choose a value:
\(\begin{align}&x_{1}=0\\&x_{2}=\frac{4}{5}\\&x_{3}=2\end{align}\)
- step7: Test the chosen value:
\(\begin{align}&x<\frac{3}{5}\textrm{ }\textrm{is not a solution}\\&\frac{3}{5}1\textrm{ }\textrm{is not a solution}\end{align}\)
- step8: The final solution is \(\frac{3}{5}\frac{\sqrt{97}-7}{8}\end{align}\)
- step12: Choose a value:
\(\begin{align}&x_{1}=-3\\&x_{2}=-1\\&x_{3}=1\end{align}\)
- step13: Test the chosen value:
\(\begin{align}&x<-\frac{\sqrt{97}+7}{8}\textrm{ }\textrm{is the solution}\\&-\frac{\sqrt{97}+7}{8}\frac{\sqrt{97}-7}{8}\textrm{ }\textrm{is the solution}\end{align}\)
- step14: Include the critical value:
\(\begin{align}&x\leq -\frac{\sqrt{97}+7}{8}\textrm{ }\textrm{is the solution}\\&x\geq \frac{\sqrt{97}-7}{8}\textrm{ }\textrm{is the solution}\end{align}\)
- step15: The final solution is \(x \in \left(-\infty,-\frac{\sqrt{97}+7}{8}\right]\cup \left[\frac{\sqrt{97}-7}{8},+\infty\right):\)
\(x \in \left(-\infty,-\frac{\sqrt{97}+7}{8}\right]\cup \left[\frac{\sqrt{97}-7}{8},+\infty\right)\)
Solve the equation \( 2 x^{2}+3 x-4 \leq 0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(2x^{2}+3x-4\leq 0\)
- step1: Rewrite the expression:
\(2x^{2}+3x-4=0\)
- step2: Add or subtract both sides:
\(2x^{2}+3x=4\)
- step3: Divide both sides:
\(\frac{2x^{2}+3x}{2}=\frac{4}{2}\)
- step4: Evaluate:
\(x^{2}+\frac{3}{2}x=2\)
- step5: Add the same value to both sides:
\(x^{2}+\frac{3}{2}x+\frac{9}{16}=2+\frac{9}{16}\)
- step6: Simplify the expression:
\(\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}\)
- step7: Simplify the expression:
\(x+\frac{3}{4}=\pm \sqrt{\frac{41}{16}}\)
- step8: Simplify the expression:
\(x+\frac{3}{4}=\pm \frac{\sqrt{41}}{4}\)
- step9: Separate into possible cases:
\(\begin{align}&x+\frac{3}{4}=\frac{\sqrt{41}}{4}\\&x+\frac{3}{4}=-\frac{\sqrt{41}}{4}\end{align}\)
- step10: Solve the equation:
\(\begin{align}&x=\frac{\sqrt{41}-3}{4}\\&x=-\frac{\sqrt{41}+3}{4}\end{align}\)
- step11: Determine the test intervals:
\(\begin{align}&x<-\frac{\sqrt{41}+3}{4}\\&-\frac{\sqrt{41}+3}{4}\frac{\sqrt{41}-3}{4}\end{align}\)
- step12: Choose a value:
\(\begin{align}&x_{1}=-3\\&x_{2}=-1\\&x_{3}=2\end{align}\)
- step13: Test the chosen value:
\(\begin{align}&x<-\frac{\sqrt{41}+3}{4}\textrm{ }\textrm{is not a solution}\\&-\frac{\sqrt{41}+3}{4}\frac{\sqrt{41}-3}{4}\textrm{ }\textrm{is not a solution}\end{align}\)
- step14: Include the critical value:
\(\begin{align}&-\frac{\sqrt{41}+3}{4}\leq x\leq \frac{\sqrt{41}-3}{4}\textrm{ }\textrm{is the solution}\end{align}\)
- step15: The final solution is \(-\frac{\sqrt{41}+3}{4}\leq x\leq \frac{\sqrt{41}-3}{4}:\)
\(-\frac{\sqrt{41}+3}{4}\leq x\leq \frac{\sqrt{41}-3}{4}\)
Solve the equation \( x^{2}+3 x+2<0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(x^{2}+3x+2<0\)
- step1: Rewrite the expression:
\(x^{2}+3x+2=0\)
- step2: Factor the expression:
\(\left(x+1\right)\left(x+2\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&x+1=0\\&x+2=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=-1\\&x=-2\end{align}\)
- step5: Determine the test intervals:
\(\begin{align}&x<-2\\&-2-1\end{align}\)
- step6: Choose a value:
\(\begin{align}&x_{1}=-3\\&x_{2}=-\frac{3}{2}\\&x_{3}=0\end{align}\)
- step7: Test the chosen value:
\(\begin{align}&x<-2\textrm{ }\textrm{is not a solution}\\&-2-1\textrm{ }\textrm{is not a solution}\end{align}\)
- step8: The final solution is \(-20 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(3x^{2}-9x+6>0\)
- step1: Rewrite the expression:
\(3x^{2}-9x+6=0\)
- step2: Factor the expression:
\(3\left(x-2\right)\left(x-1\right)=0\)
- step3: Divide the terms:
\(\left(x-2\right)\left(x-1\right)=0\)
- step4: Separate into possible cases:
\(\begin{align}&x-2=0\\&x-1=0\end{align}\)
- step5: Solve the equation:
\(\begin{align}&x=2\\&x=1\end{align}\)
- step6: Determine the test intervals:
\(\begin{align}&x<1\\&12\end{align}\)
- step7: Choose a value:
\(\begin{align}&x_{1}=0\\&x_{2}=\frac{3}{2}\\&x_{3}=3\end{align}\)
- step8: Test the chosen value:
\(\begin{align}&x<1\textrm{ }\textrm{is the solution}\\&12\textrm{ }\textrm{is the solution}\end{align}\)
- step9: The final solution is \(x \in \left(-\infty,1\right)\cup \left(2,+\infty\right):\)
\(x \in \left(-\infty,1\right)\cup \left(2,+\infty\right)\)
Aquí están las soluciones para las inecuaciones que has planteado:
a. Para la inecuación \( x^{2}+3 x+2<0 \):
- La solución es: \(-2 < x < -1\)
b. Para la inecuación \( 3 x^{2}-9 x+6>0 \):
- La solución es: \( x \in (-\infty, 1) \cup (2, +\infty) \)
c. Para la inecuación \( 5 x^{2}-8 x+3<0 \):
- La solución es: \(\frac{3}{5} < x < 1\)
d. Para la inecuación \( 8 x^{2}+14 x<-6 \):
- La solución es: \(-1 < x < -\frac{3}{4}\)
e. Para la inecuación \( 2 x^{2}+3 x-4 \leq 0 \):
- La solución es: \(-\frac{\sqrt{41}+3}{4} \leq x \leq \frac{\sqrt{41}-3}{4}\)
f. Para la inecuación \( 4 x^{2}+7 x-3 \geq 0 \):
- La solución es: \( x \in \left(-\infty, -\frac{\sqrt{97}+7}{8}\right) \cup \left[\frac{\sqrt{97}-7}{8}, +\infty\right) \)
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