Lesson 7

What do you notice? What do you wonder?

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

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.

   Figure A

   

   

   Figure B

   

   

   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?

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 \((x+3)^2 = 4\)?

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.

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. \(\sqrt{z} + 9 = 0\)

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!