To solve the equation [tex]\(2^x + \frac{1}{2^x} = 4 \frac{1}{4}\)[/tex], let's proceed step-by-step.
1. Rewrite the Right-Hand Side:
- Notice that [tex]\(4 \frac{1}{4}\)[/tex] can also be written as [tex]\(4.25\)[/tex]. This makes the equation:
[tex]\[
2^x + \frac{1}{2^x} = 4.25
\][/tex]
2. Introduce a New Variable:
- Let [tex]\(y = 2^x\)[/tex]. This transforms the equation to:
[tex]\[
y + \frac{1}{y} = 4.25
\][/tex]
3. Multiply Both Sides by y:
- To get rid of the fraction, multiply both sides by [tex]\(y\)[/tex]:
[tex]\[
y^2 + 1 = 4.25y
\][/tex]
4. Rearrange into a Standard Quadratic Equation:
- Bring all terms to one side to form a quadratic equation:
[tex]\[
y^2 - 4.25y + 1 = 0
\][/tex]
5. Solve the Quadratic Equation:
- We use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4.25\)[/tex], and [tex]\(c = 1\)[/tex].
[tex]\[
y = \frac{-(-4.25) \pm \sqrt{(-4.25)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{4.25 \pm \sqrt{18.0625 - 4}}{2}
\][/tex]
[tex]\[
y = \frac{4.25 \pm \sqrt{14.0625}}{2}
\][/tex]
[tex]\[
y = \frac{4.25 \pm 3.75}{2}
\][/tex]
6. Find the Roots:
- Solve for the two possible values of [tex]\(y\)[/tex]:
[tex]\[
y = \frac{4.25 + 3.75}{2} \quad \text{and} \quad y = \frac{4.25 - 3.75}{2}
\][/tex]
[tex]\[
y = \frac{8}{2} = 4 \quad \text{and} \quad y = \frac{0.5}{2} = 0.25
\][/tex]
Thus, the solutions for [tex]\(y\)[/tex] are:
[tex]\[
y = 4 \quad \text{or} \quad y = 0.25
\][/tex]
7. Return to the Original Variable [tex]\(x\)[/tex]:
- Recall that [tex]\(y = 2^x\)[/tex]. We need to solve for [tex]\(x\)[/tex] in each of the cases [tex]\(2^x = 4\)[/tex] and [tex]\(2^x = 0.25\)[/tex].
[tex]\[
2^x = 4 \quad \Rightarrow \quad 2^x = 2^2 \quad \Rightarrow \quad x = 2
\][/tex]
[tex]\[
2^x = 0.25 \quad \Rightarrow \quad 2^x = 2^{-2} \quad \Rightarrow \quad x = -2
\][/tex]
Therefore, the solutions to the equation [tex]\(2^x + \frac{1}{2^x} = 4.25\)[/tex] are:
[tex]\[
x = -2 \quad \text{and} \quad x = 2
\][/tex]
