# Lesson 7

Inequivalent Equations

- Let’s see what happens when we square each side of an equation.

### 7.1: 2 and -2

What do you notice? What do you wonder?

- \(x^2 = 4\)
- \(x^2 = \text-4\)
- \((x-2)(x+2)=0\)
- \(x=\sqrt{4}\)

### 7.2: Careful When You Take the Square Root

Tyler was solving this equation:

\(\displaystyle x^2 - 1 = 3\)

He said, “I can add 1 to each side of the equation and it doesn’t change the equation. I get \(x^2 = 4\).”

- Priya said, “It does change the equation. It just doesn’t change the solutions!” Then she showed these two graphs.
- How can you see the solution to the equation \(x^2 - 1 = 3\) in Figure A?
- How can you see the solution to the equation \(x^2 = 4\) in Figure B?
- Use the graphs to explain why the equations have the same solutions.

- Tyler said, “Now I can take the square root of each side to get the solution to \(x^2 = 4\). The square root of \(x^2\) is \(x\). The square root of 4 is 2.” He wrote:
\(\begin{align}x^2 &= 4\\ \sqrt{x^2} &= \sqrt{4} \\ x &= 2 \end{align}\)

Priya said, “But the graphs show that there are

*two*solutions!” What went wrong?

### 7.3: Another Way to Solve

Han was solving this equation: \(\displaystyle \dfrac{x+3}{2} = 4\)

He said, "I know that half of \(x+3\) is 4. So \(x+3\) must be 8, since half of 8 is 4. This means that \(x\) is 5."

- Use Han's reasoning to solve this equation: \((x+3)^2 = 4\).
- What advice would you give to someone who was going to solve an equation like \((x+3)^2 = 4\)?

### 7.4: What Happens When You Square Each Side?

Mai was solving this equation: \(\displaystyle \sqrt{x-1} = 3\)

She said, “I can square each side of the equation to get another equation with the same solutions.” Then she wrote:

\(\begin{align}\sqrt{x-1} &= 3\\ (\sqrt{x-1})^2 &= 3^2 \\ x-1 &= 9 \\ x &= 10 \end{align}\)

- Check to see if her solution makes the original equation true.
- Andre said, “I tried your technique to solve \(\displaystyle \sqrt{x-1} = \text{-} 3\)but it didn’t work.” Why doesn’t it work? Explain or show your reasoning.

### 7.5: Solve These Equations With Square Roots in Them

Find the solution(s) to each of these equations, or explain why there is no solution.

- \(\sqrt{t + 4} = 3\)
- \(\text{-}10 = \text{-}\sqrt{a}\)
- \(\sqrt{3 - w} - 4 = 0\)
- \(\sqrt{z} + 9 = 0\)

Are there values of \(a\) and \(b\) so that the equation \(\sqrt{t + a} = b\) has more than one solution? Explain your reasoning.

### Summary

Every positive number has *two* square roots. You can see this by looking at the graph of \(y=x^2\):

When we have a square root in an equation like \(\sqrt{t} - 6 = 0\), we can isolate the square root and then square each side:

\(\begin{align}\sqrt{t} - 6 &= 0\\ \sqrt{t} &= 6\\ t &= 6^2 \\ t &= 36 \end{align}\)

But sometimes, squaring each side of an equation gives results that aren’t solutions to the original equation. For example:

\(\begin{align}\sqrt{t} + 6 &= 0\\ \sqrt{t} &= \text-6\\ t &= (\text-6)^2 \\ t &= 36 \end{align}\)

Note that 36 is *not* a solution to the original equation, because \(\sqrt{36}+6\) doesn’t equal 0. In fact, \(\sqrt{t}+6=0\) has no solutions, because it’s impossible for the sum of two positive numbers to be zero.

Remember: sometimes the new equation has solutions that the old equation doesn’t have. Always check your solutions in the original equation!