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Chapter 6: Problem 74

Solve each equation, and check your solutions. $$ 5-(x-1)^{2}=(x-2)^{2} $$

Short Answer

Expert verified
The solutions are \( x = 0 \) and \( x = 3 \).

Step by step solution

01

- Expand both sides

First, expand the equation \[ 5 - (x-1)^{2} = (x-2)^{2} \] The equation becomes: \[ 5 - (x^2 - 2x + 1) = x^2 - 4x + 4 \] This simplifies to: \[ 5 - x^2 + 2x - 1 = x^2 - 4x + 4 \] which is: \[ 4 - x^2 + 2x = x^2 - 4x + 4 \]
02

- Combine like terms

Bring all the terms to one side to set the equation to 0: \[ 4 - x^2 + 2x - x^2 + 4x - 4 = 0 \] Combining like terms: \[ -2x^2 + 6x = 0 \]
03

- Solve the quadratic equation

Factor out the common term: \[ -2x(x - 3) = 0 \] This gives two solutions: \[ x = 0 \] and \[ x = 3 \]
04

- Check the solutions

Substitute the solutions back into the original equation to verify. For \( x = 0 \): \[ 5 - (0-1)^{2} = (0-2)^{2} \] \[ 5 - 1 = 4 \] \[ 4 = 4 \] So, \( x = 0 \) is a solution. For \( x = 3 \): \[ 5 - (3-1)^{2} = (3-2)^{2} \] \[ 5 - 4 = 1 \] \[ 1 = 1 \] So, \( x = 3 \) is a solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expanding Equations
In the given problem, we start by expanding both sides of the equation. Expanding equations allows us to remove parentheses by distributing and re-writing expressions. For example, the equation given is: 5 - (x-1)^{2} = (x-2)^{2}.
First, we expand (x-1)^{2} on the left side and (x-2)^{2} on the right side. Remember that expanding means multiplying the binomials by themselves:
(x-1)^{2} becomes: (x-1)(x-1) = x^2 - x - x + 1 = x^2 - 2x + 1.
Similarly, (x-2)^{2} expands to: (x-2)(x-2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4.
So, the expanded version of the given equation is: 5 - (x^2 - 2x + 1) = x^2 - 4x + 4.
Combining Like Terms
After expanding, our next step is to combine like terms. Like terms are terms that have the same variable raised to the same power. By combining them, we simplify the equation further.
For our equation: 5 - x^2 + 2x - 1 = x^2 - 4x + 4.
Notice that we have terms with x^2, x, and constant terms (numbers without variables).
We can rewrite this as: 4 - x^2 + 2x = x^2 - 4x + 4.
To combine like terms more effectively, we bring all terms to one side of the equation: 4 - x^2 + 2x - x^2 + 4x - 4 = 0.
Simplify by combining the terms: -2x^2 + 6x = 0. This step gets us closer to solving the quadratic equation.
Solving Quadratic Equations
Now, we move on to solving the quadratic equation -2x^2 + 6x = 0.
A quadratic equation is any equation that can be rearranged in the standard form: ax^2 + bx + c = 0. In our case, a = -2, b = 6, and c = 0.
First, factor out the common term, which is -2x: -2x(x - 3) = 0. This gives us two possible solutions: x = 0 and x = 3.
To verify, we substitute these solutions back into the original equation: For x = 0: 5 - (0-1)^{2} = (0-2)^{2} 5 - 1 = 4 4 = 4, which is true. For x = 3: 5 - (3-1)^{2} = (3-2)^{2} 5 - 4 = 1 1 = 1, which is true. Both solutions satisfy the original equation!

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