Certainly! Let's solve the given equation step by step.
We start with the equation:
[tex]\[
\frac{2x^2 + 1}{2x^2} = 2
\][/tex]
1. Simplify the left side of the equation:
We can separate the fraction into two parts:
[tex]\[
\frac{2x^2 + 1}{2x^2} = \frac{2x^2}{2x^2} + \frac{1}{2x^2}
\][/tex]
Simplifying further:
[tex]\[
\frac{2x^2}{2x^2} + \frac{1}{2x^2} = 1 + \frac{1}{2x^2}
\][/tex]
2. Set up the simplified equation:
Now we have:
[tex]\[
1 + \frac{1}{2x^2} = 2
\][/tex]
3. Isolate the fraction term:
Subtract 1 from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[
\frac{1}{2x^2} = 2 - 1
\][/tex]
Simplifying:
[tex]\[
\frac{1}{2x^2} = 1
\][/tex]
4. Solve for [tex]\(x^2\)[/tex]:
Take the reciprocal of both sides to solve for [tex]\(2x^2\)[/tex]:
[tex]\[
2x^2 = 1
\][/tex]
5. Isolate [tex]\(x^2\)[/tex]:
Divide both sides by 2:
[tex]\[
x^2 = \frac{1}{2}
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Take the square root of both sides to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \pm \sqrt{\frac{1}{2}}
\][/tex]
Which can be simplified:
[tex]\[
x = \pm \frac{\sqrt{2}}{2}
\][/tex]
Therefore, the solutions to the equation [tex]\(\frac{2x^2 + 1}{2x^2} = 2\)[/tex] are:
[tex]\[
x = -\frac{\sqrt{2}}{2} \quad \text{and} \quad x = \frac{\sqrt{2}}{2}
\][/tex]
