# 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 $$x^2 = 4$$.”

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

1. Use Han's reasoning to solve this equation: $$(x+3)^2 = 4$$.
2. 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}

1. Check to see if her solution makes the original equation true.
2. 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.

1. $$\sqrt{t + 4} = 3$$
2. $$\text{-}10 = \text{-}\sqrt{a}$$
3. $$\sqrt{3 - w} - 4 = 0$$
4. $$\sqrt{z} + 9 = 0$$

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 $$y=x^2$$:

If $$y$$ is a positive number like 4, then we can see that $$y=4$$ crosses the graph in two places, so the equation $$x^2=4$$ will have two solutions, namely, $$\sqrt{4}$$ and $$\text{-}\sqrt{4}$$. This is true for any positive number $$a$$: $$y = a$$ will cross the graph in two places, and $$x^2 = a$$ will have two solutions, $$x= \sqrt{a}$$ and $$x = \text{-}\sqrt{a}$$.

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!