Problem 70 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(3-2(x-1)^{-1}=(x-1)^{-2}\)

Step-by-Step Solution

Answer

The solutions are \( x = 2 \) and \( x = \frac{2}{3} \)

Step 1: Simplify the Exponents

Rewrite the equation with all terms on one side. We start with: \[ 3 - 2(x-1)^{-1} = (x-1)^{-2} \] Multiply by \( (x-1)^{-2} \) to get rid of the negative exponents: \[ 3(x-1)^2 - 2(x-1) = 1 \]

Step 2: Expand and Simplify

Expand \[ 3(x-1)^2 \] to \[ 3(x^2 - 2x + 1) \]: \[ 3(x^2 - 2x + 1) - 2(x - 1) = 1 \] Distribute 3 and -2: \[ 3x^2 - 6x + 3 - 2x + 2 = 1 \]

Step 3: Combine Like Terms

Combine like terms to simplify the equation: \[ 3x^2 - 8x + 5 = 1 \] Move 1 to the other side to set the equation to zero: \[ 3x^2 - 8x + 4 = 0 \]

Step 4: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = -8 \), and \( c = 4 \). Substitute these values into the formula: \[ x = \frac{8 \pm \sqrt{64 - 48}}{6} \] Simplify under the square root and solution: \[ x = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6} \] This gives: \[ x = 2 \] or \[ x = \frac{2}{3} \]

Step 5: Check the Solutions

Verify each solution by substituting back into the original equation. For \( x = 2 \): \[ 3 - 2(2-1)^{-1} = (2-1)^{-2} \] simplifies to \[ 3 - 2 = 1 \] which is correct. Now check \( x = \frac{2}{3} \): \[ 3 - 2\left(\frac{2}{3} - 1\right)^{-1} = \left(\frac{2}{3} - 1\right)^{-2} \] simplifying to \[ 3 - 2(-\frac{1}{3})^{-1} = (-\frac{1}{3})^{-2} \] This leads to \[ 3 - 2(-3) = 9 \] which simplifies to \[ 3 + 6 = 9 \] which is correct.

Key Concepts

Negative Exponents

Negative exponents can be confusing at first, but with a bit of practice, they become much easier to handle. A negative exponent indicates that a number should be taken to the denominator. For instance, the expression \(x^{-n}\) becomes \(\frac{1}{x^n}\). So, in our equation \(3 - 2(x-1)^{-1} = (x-1)^{-2}\), the term \( (x-1)^{-1} \) is equivalent to \( \frac{1}{(x-1)} \) and \( (x-1)^{-2} \) equals \( \frac{1}{(x-1)^2} \).
Notice how these transformations make the equation more manageable for further steps. When you multiply both sides by the highest degree of the denominator—in this case, \( (x-1)^2 \)—you get rid of the fractions, making the equation easier to solve.

Quadratic Formula

The quadratic formula is a mathematical tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the given exercise, we simplified the original equation to the standard form \(3x^2 - 8x + 4 = 0\). Here:

• \(a = 3\)
• \(b = -8\)
• \(c = 4\)

To find the solutions, we substitute these values into the quadratic formula:
\[ x = \frac{8 \pm \sqrt{64 - 48}}{6} \]
This further simplifies to \[ \frac{8 \pm 4}{6} \], giving us \[ x = 2 \] or \[ x = \frac{2}{3} \].
Always remember to substitute the values correctly and simplify each part of the formula step by step. Double-checking each step helps avoid mistakes.

Checking Solutions

After finding the solutions to the equation, it is crucial to check if they actually satisfy the original equation. Sometimes, transformations can introduce extraneous solutions. This means that not all solutions obtained from the quadratic formula are valid.
For our case:

• **For \( x = 2 \):
  Substitute back into the original equation:
  \[ 3 - 2(2-1)^{-1} = (2-1)^{-2} \]
  simplifies to \[ 3 - 2 = 1 \], which is true.


• **For \( x = \frac{2}{3} \):
  Substitute into the equation:
  \[ 3 - 2\left(\frac{2}{3} - 1\right)^{-1} = \left(\frac{2}{3} - 1\right)^{-2} \]
  simplifies to \[ 3 - 2\left(-\frac{1}{3}\right)^{-1} = \left(-\frac{1}{3}\right)^{-2} \], which leads to \[ 3 - 2(-3) = 9 \], and so \[ 3 + 6 = 9 \], which is also correct.

Checking your solutions not only verifies their correctness but also deepens your understanding of how different transformations affect the equation.