To solve the equation [tex]\(2^x + \frac{1}{2^x} = 4 + \frac{1}{4}\)[/tex], follow these detailed steps:
1. Simplify the right-hand side of the equation:
[tex]\[
4 + \frac{1}{4}
\][/tex]
Convert the mixed number into a single fraction or a decimal to simplify computation:
[tex]\[
4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}
\][/tex]
So, the equation now looks like:
[tex]\[
2^x + \frac{1}{2^x} = \frac{17}{4}
\][/tex]
2. Multiply both sides of the equation by [tex]\(2^x\)[/tex] to clear the fraction on the left-hand side:
[tex]\[
2^x \left(2^x + \frac{1}{2^x}\right) = 2^x \cdot \frac{17}{4}
\][/tex]
This simplifies to:
[tex]\[
(2^x)^2 + 1 = \frac{17 \cdot 2^x}{4}
\][/tex]
[tex]\[
2^{2x} + 1 = \frac{17 \cdot 2^x}{4}
\][/tex]
3. Rearrange the equation to form a quadratic in terms of [tex]\(2^x\)[/tex]:
Let [tex]\(y = 2^x\)[/tex]. The equation then becomes:
[tex]\[
y^2 + 1 = \frac{17y}{4}
\][/tex]
4. Clear the fraction by multiplying through by 4:
[tex]\[
4y^2 + 4 = 17y
\][/tex]
5. Rearrange the terms to set the equation to zero:
[tex]\[
4y^2 - 17y + 4 = 0
\][/tex]
6. Solve the quadratic equation [tex]\(4y^2 - 17y + 4 = 0\)[/tex]:
Use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -17\)[/tex], and [tex]\(c = 4\)[/tex]:
[tex]\[
y = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4}
\][/tex]
[tex]\[
y = \frac{17 \pm \sqrt{289 - 64}}{8}
\][/tex]
[tex]\[
y = \frac{17 \pm \sqrt{225}}{8}
\][/tex]
[tex]\[
y = \frac{17 \pm 15}{8}
\][/tex]
This results in two possible values for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{17 + 15}{8} = \frac{32}{8} = 4
\][/tex]
[tex]\[
y = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4}
\][/tex]
7. Back substitute [tex]\(y = 2^x\)[/tex] to find [tex]\(x\)[/tex]:
- For [tex]\(y = 4\)[/tex]:
[tex]\[
2^x = 4 \implies 2^x = 2^2 \implies x = 2
\][/tex]
- For [tex]\(y = \frac{1}{4}\)[/tex]:
[tex]\[
2^x = \frac{1}{4} \implies 2^x = 2^{-2} \implies x = -2
\][/tex]
Thus, the solutions to the equation [tex]\(2^x + \frac{1}{2^x} = 4 + \frac{1}{4}\)[/tex] are [tex]\(x = 2\)[/tex] and [tex]\(x = -2\)[/tex].
