### 7.5.4 Multiple Solutions

In this section we’ll explore equations with more than one solution. We’ll be using a very common problem-solving strategy: if you want to solve a complicated problem, invent a related problem that’s much easier, and solve that first to get some ideas. If we want to solve an equation like $\abs{2x-3} + 5 = \abs{x^2-7}$ we might start with something simple like $\abs{x}=5$ and work up from there.

#### Zero Product

It’s common to encounter an equation that has some expressions multiplied together equal to zero, for example: $3x(1-2x)(x^2+5) = 0.$ Let’s start with a very simple version.

##### Question 7.5.10 What does the equation $xy=0$ tell us about $x$ and $y$? What are some possible values of $x$ and $y$ that solve the equation? Show hint

##### Question 7.5.11 Find all solutions to the equation $(x+2)(x-5) = 0$. Show hint

When there is more than one solution, for example if $x=1$ and $x=2$ will both satisfy an equation, then we usually write them separated by a comma, like this: $x=1,2$ meaning $x$ is $1$ or $2$.

When we have an equation like $(x+2)(x-4)=0$ for example, we need to split it up into two cases: either $x+2=0$ or $x-4=0$. I’ll show you some ways to neatly and clearly set this out.

Some people like to write it like this: $(x+2)(x-4) = 0$ \begin{alignat*}{2} &\text{Case }1:\quad&x+2 &= 0\\ &&x &= -2\\\\ &\text{Case }2:\quad&x-4 &= 0\\ &&x &= 4 \end{alignat*} $\therefore x=-2,4.$

I prefer to put the different cases in vertical columns like this: \begin{gather*} (x+2)(x-4) = 0\\\\ \begin{aligned}[t] x+2 &= 0\\ x &= -2 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] x-4 &= 0\\ x &= 4 \end{aligned}\\\\ \therefore x=-2,4. \end{gather*} This is the way I’ll be writing solutions out in this section, but you can choose whichever way you like.

##### Question 7.5.12 Find all solutions to the equation $x(3x+2)(x-4) = 0$ Show answer

Remember:
When two or more expressions multiplied together equal zero, consider a case for each one being zero.

#### Cancelling

Multiple solutions can also arise when we want to cancel a common factor from both sides of an equation, but we’re not sure that it’s non-zero.

##### Question 7.5.13 Find all solutions to the equation $x^2=3x$. Show hint

##### Question 7.5.14 Solve $5k(k+7) = 2(k+7)$ for $k$. Show hint

##### Question 7.5.15 Solve $2(x-3a) = (x-3a)(x+1)$ for $x$. Show answer

Remember:
When cancelling a common factor from both sides of an equation, consider that it might be zero.

#### Squares

##### Question 7.5.16 What values can $x$ take if $x^2 = 4$? Show hint

##### Question 7.5.17 Solve $(x-3)^2 = 25$ for $x$. Show answer

##### Question 7.5.18 Solve $(2t+3)^2 + 1 = 50$ for $t$. Show answer

#### Absolute Values

##### Question 7.5.19 Solve $\abs{x} = 5$ for $x$. Show answer

##### Question 7.5.20 Find all possible values of $y$ if $\abs{y+3} = 1$. Show answer

##### Question 7.5.21 Solve $\abs{x-3} = 2x$ for $x$. Show hint

##### Question 7.5.22 Solve the equation $\abs{t-2} = \abs{3t+4}$ for $t$. Show answer

Remember:
An equation involving absolute values can be split into different cases depending on sign, but watch out for extraneous solutions.