Problem 49 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ 2^{x}+2^{-x}=2 $$

Step-by-Step Solution

Answer

The solution is \(x = 0\).

Step 1: Multiply both sides by a common factor

Begin by multiplying both sides of the equation by \(2^x\) to eliminate the fraction involving \(2^{-x}\):\[2^x imes 2^x + 2^x imes 2^{-x} = 2 imes 2^x\]This simplifies to:\[(2^x)^2 + 1 = 2^{x+1}\]

Step 2: Substitute variable

Let \(y = 2^x\). Therefore, the equation \((2^x)^2 + 1 = 2^{x+1}\) becomes:\[y^2 + 1 = 2y\]

Step 3: Rearrange into a quadratic equation

Rearrange the equation obtained in Step 2 into standard quadratic form:\[y^2 - 2y + 1 = 0\]

Step 4: Factor the quadratic equation

Notice that the quadratic equation \(y^2 - 2y + 1 = 0\) is a perfect square trinomial. Factor it as:\[(y - 1)^2 = 0\]

Step 5: Solve the factored equation

Set the factored equation \((y - 1)^2 = 0\) equal to zero and solve for \(y\):\[y - 1 = 0 \rightarrow y = 1\]

Step 6: Substitute back to solve for original variable

Recall that \(y = 2^x\). Substitute back to find \(x\):\[2^x = 1\]Since \(2^0 = 1\), the solution is:\[x = 0\]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is written as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this exercise, the quadratic you encounter is \( y^2 - 2y + 1 = 0 \).
This is derived from substituting \( y = 2^x \), which transforms the exponential equation into a quadratic equation.
Here, all terms are set in standard quadratic form, where \( a = 1 \), \( b = -2 \), and \( c = 1 \).
For such equations, the discriminant \( b^2 - 4ac \) indicates the nature of the roots. In our case, \( (-2)^2 - 4(1)(1) = 0 \), which means a perfect square trinomial.
This indicates a single, repeated root, showing the special nature of your solution when you are solving this equation using the quadratic formula.
While the quadratic formula suits any form, some quadratics like this one are easier handled by factoring. The formula, however, guarantees you can find the root, especially when factoring seems non-intuitive.

Factoring

Factoring involves breaking down a quadratic equation into products of simpler binomials.
The equation \( y^2 - 2y + 1 = 0 \) can be recognized as a perfect square trinomial.
Perfect square trinomials have the form \( (a-b)^2 = a^2 - 2ab + b^2 \), or more specifically like here, \( (y - 1)^2 = 0 \).

• This means the trinomial results from squaring a binomial \( (y - 1) \).

When you factor this trinomial, you are checking for identical terms binned within the equation.Factoring is a method based on observing patterns and identifying squares within the quadratic term.
Once factored, setting each factor to zero gives the potential solutions, being essential to finding the solutions of quadratics in simpler cases.This intuitive method makes addressing equations like \( y^2 - 2y + 1 = 0 \) straightforward, reaching \( y = 1 \) by directly utilizing algebraic identities inherent in perfect squares.

Substitution Method

The substitution method is a clever strategy when dealing with complicated exponential equations.
The idea is to simplify the equation by substituting a new variable in place of an exponential term.In this exercise, we use \( y = 2^x \) to transform a complex exponential expression into a more manageable quadratic form:

• Original Equation: \( 2^{x} + 2^{-x} = 2 \)
• Transformed Equation: \( y^2 + 1 = 2y \) by letting \( y = 2^x \)

The substitution helps convert the exponential equation to a format that is easier to solve using quadratic methods.Substitution relies on the consistency of equality: by ensuring that all transformations are algebraically identical,
you maintain the integrity of the original equation.
This helps convert otherwise challenging or unwieldy problems into quadratics that are familiar ground for those acquainted with algebra. This technique is especially useful when exponential terms can be isolated or transformed into squared terms, allowing for further simplification or factoring.