Ward O'Quinn
05/03/2023 · Primary School
Solve the inequality and write the solution in interval notation. Leave answers in exact, simplified form. \[ 2 \mathrm{x}^{2}+4 \leq 8 \mathrm{x} \] Preview
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Step-by-step Solution
Solve the equation \( 2x^2+4\leq 8x \).
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(2x^{2}+4\leq 8x\)
- step1: Move the expression to the left side:
\(2x^{2}+4-8x\leq 0\)
- step2: Rewrite the expression:
\(2x^{2}+4-8x=0\)
- step3: Add or subtract both sides:
\(2x^{2}-8x=-4\)
- step4: Divide both sides:
\(\frac{2x^{2}-8x}{2}=\frac{-4}{2}\)
- step5: Evaluate:
\(x^{2}-4x=-2\)
- step6: Add the same value to both sides:
\(x^{2}-4x+4=-2+4\)
- step7: Simplify the expression:
\(\left(x-2\right)^{2}=2\)
- step8: Simplify the expression:
\(x-2=\pm \sqrt{2}\)
- step9: Separate into possible cases:
\(\begin{align}&x-2=\sqrt{2}\\&x-2=-\sqrt{2}\end{align}\)
- step10: Move the constant to the right side:
\(\begin{align}&x=\sqrt{2}+2\\&x=-\sqrt{2}+2\end{align}\)
- step11: Determine the test intervals:
\(\begin{align}&x<-\sqrt{2}+2\\&-\sqrt{2}+2\sqrt{2}+2\end{align}\)
- step12: Choose a value:
\(\begin{align}&x_{1}=0\\&x_{2}=2\\&x_{3}=4\end{align}\)
- step13: Test the chosen value:
\(\begin{align}&x<-\sqrt{2}+2\textrm{ }\textrm{is not a solution}\\&-\sqrt{2}+2\sqrt{2}+2\textrm{ }\textrm{is not a solution}\end{align}\)
- step14: Include the critical value:
\(\begin{align}&-\sqrt{2}+2\leq x\leq \sqrt{2}+2\textrm{ }\textrm{is the solution}\end{align}\)
- step15: The final solution is \(-\sqrt{2}+2\leq x\leq \sqrt{2}+2:\)
\(-\sqrt{2}+2\leq x\leq \sqrt{2}+2\)
The solution to the inequality \(2x^2+4\leq 8x\) is \(-\sqrt{2}+2\leq x\leq \sqrt{2}+2\).
Therefore, the solution in interval notation is \([-\sqrt{2}+2, \sqrt{2}+2]\).
Quick Answer
The solution in interval notation is \([-\sqrt{2}+2, \sqrt{2}+2]\).
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