Problem 91 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$16 x^{-4}-65 x^{-2}+4=0$$

Step-by-Step Solution

Answer

x = \pm\frac{1}{2}, \pm 4

Step 1: Substitute a New Variable

Let us make the substitution to simplify the equation. Let \( u = x^{-2} \). Therefore, \( u^2 = x^{-4} \). The equation now becomes: \[ 16u^2 - 65u + 4 = 0 \]

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \(16u^2 - 65u + 4 = 0\) using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 16\), \(b = -65\), and \(c = 4\). Substituting these values in: \[ u = \frac{65 \pm \sqrt{4225 - 256}}{32} \] \[ u = \frac{65 \pm \sqrt{3969}}{32} \] \[ u = \frac{65 \pm 63}{32} \] This results in two solutions for \(u\): \[ u = \frac{65 + 63}{32} = 4 \] \[ u = \frac{65 - 63}{32} = \frac{1}{16} \]

Step 3: Re-substitute the Original Variable

Replace \( u \) back with \( x^{-2} \). So, we obtain two equations: \[ x^{-2} = 4 \] \[ x^{-2} = \frac{1}{16} \]

Step 4: Solve for x

Solve for \(x\) from the two equations. For \( x^{-2} = 4 \): \[ x^{-2} = 4 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm\frac{1}{2} \] For \( x^{-2} = \frac{1}{16} \): \[ x^{-2} = \frac{1}{16} \Rightarrow x^2 = 16 \Rightarrow x = \pm 4 \]

Key Concepts

headline of the respective core concept

Quadratic equations appear frequently in mathematics. In this exercise, we encounter a quadratic equation after substituting a variable. Quadratic equations are equations of the form:
= constant.
A general quadratic equation looks like this:
16u^2 - 65u + 4 = 0
One key feature of quadratic equations is that they are always parabolas when graphed. They can have zero, one, or two real roots, depending on the discriminant, which we will introduce later.

headline of the respective core concept

Here, we use variable substitution to make our polynomial easier to work with. Substituting a new variable can simplify complex polynomial equations into a more manageable form.
In our exercise, we let u = x^{-2}. This substitution makes the original equation:
\(16 x^{-4}-65 x^{-2}+4=0\)
become: \(16u^2 - 65u + 4 = 0\).
This step is crucial as it allows us to deal with a quadratic equation rather than a polynomial with negative exponents.

headline of the respective core concept

Negative exponents often seem intimidating, but they follow straightforward rules.
A negative exponent means you take the reciprocal of the base and then raise it to the corresponding positive exponent:
a^{-n}=1/a^n.
Example: x^{-2}=1/x^2.
Applying this, when we substitute \(u=x^{-2}\), we transform the original problem into one with positive exponents after solving \(u\). Understanding negative exponents was key to solving step 3 and transforming back to the original variable.

headline of the respective core concept

To solve any quadratic equation, you can use the quadratic formula:
x=frac{-b+-sqrt(b^2 - 4ac))/2a
Here, 'a' is the coefficient of \(u^2\), 'b' is the coefficient of 'u', and 'c' is the constant term. In our case, a = 16, b = -65, and c = 4.
We substitute these into the quadratic formula to find u:
u=frac{-(-65) +- sqrt((-65)^2 - 4 * 16 * 4))/ 2 * 16
u=frac{65 +- sqrt{4225-256}}>32)
u=frac{65 +- {63}}>32).
This calculation gives us two possible values for 'u', which we then resubstitute into the original equation.

headline of the respective core concept

Algebraic manipulation is essential in solving equations.
We started by substituting u back to the original variable:
\(u = x^{-2}\).
This gave us:
x^{-2} = 4
x^{2} = 1/4
x=+-1/2
and for x^{-2}=1/16:
x^{2}=16
x=+-4
Through algebraic manipulation, we found the possible values for x. It's the part of math that helps you rearrange formulas and solve for the variable you're interested in.