Simpson Joseph
02/01/2023 · Senior High School
ESIGUALDADES CUADRATICAS \( \begin{array}{ll}\text { i. Resuelva las siguientes desigualdades } \\ \text { a) } x^{2}-6 x+8 \leq 0 & \text { b) }-2 x^{2}+1>0\end{array} \)
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Step-by-step Solution
Solve the equation \( x^{2}-6x+8 \leq 0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(x^{2}-6x+8\leq 0\)
- step1: Rewrite the expression:
\(x^{2}-6x+8=0\)
- step2: Factor the expression:
\(\left(x-4\right)\left(x-2\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&x-4=0\\&x-2=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=4\\&x=2\end{align}\)
- step5: Determine the test intervals:
\(\begin{align}&x<2\\&24\end{align}\)
- step6: Choose a value:
\(\begin{align}&x_{1}=1\\&x_{2}=3\\&x_{3}=5\end{align}\)
- step7: Test the chosen value:
\(\begin{align}&x<2\textrm{ }\textrm{is not a solution}\\&24\textrm{ }\textrm{is not a solution}\end{align}\)
- step8: Include the critical value:
\(\begin{align}&2\leq x\leq 4\textrm{ }\textrm{is the solution}\end{align}\)
- step9: The final solution is \(2\leq x\leq 4:\)
\(2\leq x\leq 4\)
Solve the equation \( -2x^{2}+1>0 \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(-2x^{2}+1>0\)
- step1: Rewrite the expression:
\(-2x^{2}+1=0\)
- step2: Move the constant to the right side:
\(-2x^{2}=0-1\)
- step3: Remove 0:
\(-2x^{2}=-1\)
- step4: Change the signs:
\(2x^{2}=1\)
- step5: Divide both sides:
\(\frac{2x^{2}}{2}=\frac{1}{2}\)
- step6: Divide the numbers:
\(x^{2}=\frac{1}{2}\)
- step7: Simplify the expression:
\(x=\pm \sqrt{\frac{1}{2}}\)
- step8: Simplify:
\(x=\pm \frac{\sqrt{2}}{2}\)
- step9: Separate into possible cases:
\(\begin{align}&x=\frac{\sqrt{2}}{2}\\&x=-\frac{\sqrt{2}}{2}\end{align}\)
- step10: Determine the test intervals:
\(\begin{align}&x<-\frac{\sqrt{2}}{2}\\&-\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}\end{align}\)
- step11: Choose a value:
\(\begin{align}&x_{1}=-2\\&x_{2}=0\\&x_{3}=2\end{align}\)
- step12: Test the chosen value:
\(\begin{align}&x<-\frac{\sqrt{2}}{2}\textrm{ }\textrm{is not a solution}\\&-\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}\textrm{ }\textrm{is not a solution}\end{align}\)
- step13: The final solution is \(-\frac{\sqrt{2}}{2}0\) es \(-\frac{\sqrt{2}}{2} < x < \frac{\sqrt{2}}{2}\).
Quick Answer
a) \(2 \leq x \leq 4\)
b) \(-\frac{\sqrt{2}}{2} < x < \frac{\sqrt{2}}{2}\)
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