Problem 66 - Precalculus Student Solutions Manual 5th

Question

Explain why the following equations have the same solution set. (Do not actually solve. \()\) $$ -2 x^{2}+3 x-6=0 \quad \text { and } \quad 2 x^{2}-3 x+6=0 $$

Step-by-Step Solution

Answer

Since the second equation is \(-1)\ times the first equation, they have the same solution set.

Step 1 Title - Write down both equations

The first equation is \( -2x^2 + 3x - 6 = 0 \) and the second equation is \( 2x^2 - 3x + 6 = 0 \).

Step 2 Title - Multiply the second equation by -1

To compare the two equations, multiply the entire second equation by -1: \(-1(2x^2 - 3x + 6) = -2x^2 + 3x - 6 \).

Step 3 Title - Observe that the equations are identical

After multiplying the second equation by -1, it becomes \(-2x^2 + 3x - 6 = 0\), which is identical to the first equation.

Step 4 Title - Conclude their solution sets are the same

Since the transformed second equation is identical to the first equation, both equations have the same solution set.

Key Concepts

Comparing Quadratic Equations

Quadratic equations often look different, but they can sometimes share the same solutions. To compare quadratic equations, follow these steps:

• Write down both equations clearly.
• Look for any noticeable differences in their terms.
• Perform operations like multiplying by a constant to see if they can be made identical.

In our example, the equations are:
\(-2x^2 + 3x - 6 = 0 \) and \( 2x^2 - 3x + 6 = 0 \). Notice that one equation can become the other by multiplying by -1. This comparison tells us that they might have similar solution sets.

Identifying Identical Equations

Identifying identical equations involves transforming one equation to see if it matches another. Here's how you can do it:

• First, write down both quadratic equations.


• Next, look for constants. These are numbers that can be used to multiply one equation to match the other.


• In this example, multiplying \(2x^2 - 3x + 6 = 0\) by -1 leverages such constants.
  This gives us:

\[-1(2x^2 - 3x + 6) = -2x^2 + 3x - 6 \]
Now, you see both equations are identical: \( -2x^2 + 3x - 6 = 0 \). Therefore, they essentially represent the same solution set.

Multiplication of Equations by Constants

Multiplying equations by constants is a useful strategy in identifying equivalent equations. When you multiply all terms of an equation by a constant, the equation's solutions remain unchanged.
Here’s a simple process:

• Select a constant to multiply (often -1 if you need to change signs).

In our problem, the second quadratic equation \(2x^2 - 3x + 6 = 0\) was multiplied by -1. This gave:
\[-1(2x^2 - 3x + 6) = -2x^2 + 3x - 6 \]
Now both equations are identical, confirming they have the same solutions.
Remember, this technique is handy for proving that two different-looking equations are practically the same!