Sure, let's solve the given equation step-by-step:
We are given the equation:
[tex]\[ 2^x + \frac{1}{2^x} = 4 \frac{1}{4} \][/tex]
First, let's rewrite [tex]\(4 \frac{1}{4}\)[/tex] as a decimal:
[tex]\[ 4 \frac{1}{4} = 4 + \frac{1}{4} = 4 + 0.25 = 4.25 \][/tex]
So the equation becomes:
[tex]\[ 2^x + \frac{1}{2^x} = 4.25 \][/tex]
To solve this equation, we can let [tex]\( y = 2^x \)[/tex]. Then, [tex]\(\frac{1}{2^x} = \frac{1}{y}\)[/tex], and the equation becomes:
[tex]\[ y + \frac{1}{y} = 4.25 \][/tex]
Next, we need to solve this equation for [tex]\( y \)[/tex].
Multiplying both sides by [tex]\( y\)[/tex] to clear the fraction gives us:
[tex]\[ y^2 + 1 = 4.25y \][/tex]
Rearranging this into a standard quadratic equation form, we get:
[tex]\[ y^2 - 4.25y + 1 = 0 \][/tex]
Now, we solve this quadratic equation using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( y^2 - 4.25y + 1 = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = -4.25, \quad c = 1 \][/tex]
Substituting these values into the quadratic formula gives:
[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]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4.25 + 3.75}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ y = \frac{4.25 - 3.75}{2} = \frac{0.5}{2} = 0.25 \][/tex]
Recalling that [tex]\( y = 2^x \)[/tex], we now have two equations to solve for [tex]\(x\)[/tex]:
[tex]\[ 2^x = 4 \][/tex]
[tex]\[ 2^x = 0.25 \][/tex]
Solving for [tex]\( x \)[/tex] in each case:
1. For [tex]\( 2^x = 4 \)[/tex]:
[tex]\[ 4 = 2^2 \Rightarrow x = 2 \][/tex]
2. For [tex]\( 2^x = 0.25 \)[/tex]:
[tex]\[ 0.25 = \frac{1}{4} = 2^{-2} \Rightarrow x = -2 \][/tex]
Therefore, the solutions to the equation are:
[tex]\[ x = 2 \quad \text{and} \quad x = -2 \][/tex]
Hence, the values of [tex]\(x\)[/tex] that satisfy the original equation are:
[tex]\[ x = 2 \quad \text{and} \quad x = -2 \][/tex]
