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.

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.

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

Can they both be non-zero? Do they both have to be zero?

Zero times any number is zero, so if $x=0$ then $y$ can be anything and vice versa. There’s no way two non-zero numbers can multiply together to make zero, so at least one of them has to be zero, but they don’t both have to be zero. Some examples of values that would solve the equation: \begin{alignat*}{2} x&=\frac{2}{3}&\quad\text{and}\quad y&=0\\ x&=0&\quad\text{and}\quad y&=-5\\ x&=0&\quad\text{and}\quad y&=0 \end{alignat*}

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

Just like the previous question, we have two things multiplied together being equal to zero, so one of them must be zero.

Either $x+2 = 0$ or $x-5=0$. If $x+2=0$ then $x=-2$, so that is one solution. You can check by substituting it into the original equation if you like: you get $0\times3$ on the lhs, which equals zero.

The other possibility is that $x-5=0$. This gives $x=5$, the second solution.

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.

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

This time we have three cases to consider: \begin{gather*} x(3x+2)(x-4) = 0\\ x=0 \qquad\text{or}\qquad \begin{aligned}[t] 3x+2 &= 0\\ 3x &= -2\\ x &= -\frac{2}{3} \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] x-4 &= 0\\ x &= 4 \end{aligned} \end{gather*} Therefore $x=-\frac{2}{3},0,4$.

Remember:

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

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.

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

If $x$ is not zero, then we can divide both sides by it.

We’d like to divide both sides of the equation by $x$ to get $x=3$, but we can only do that if $x$ is not zero. If $x$ is zero, then each side of the equation equals zero, so $x=0$ is a solution, one which we would have missed if we’d just canceled the $x$s on both sides. So we should split it up into two cases: either $x$ is zero, in which case we’ve got a solution, or $x$ is not zero, in which case $x=3$. Therefore $x=0,3$.

Note that if the equation had been $\frac{x^2}{3x} = 1$ instead, then $x=0$ would not be a solution, because the lhs would be undefined for that value.

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

Either $k+7$ is zero, or it’s not and we can divide by it.

We’d like to divide both sides of the equation by $k+7$ to get $5k = 2$, but we can’t do that if $k+7$ is zero, so we need to consider two cases: \begin{gather*} 5k(k+7) = 2(k+7)\\ \begin{aligned}[t] k+7 &= 0\\ k &= -7 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] 5k &= 2\\ k &= \frac{2}{5} \end{aligned} \end{gather*} Therefore $k=-7,\frac{2}{5}$.

It’s also possible to rearrange the equation into $(5k-2)(k+7) = 0$ and solve it like the one in Question 7.5.12, but the first way is quicker.

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

\begin{gather*} 2(x-3a) = (x-3a)(x+1)\\ \begin{aligned}[t] x-3a &= 0\\ x &= 3a \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] 2 &= x+1\\ x &= 1 \end{aligned}\\ \therefore x=1,3a \end{gather*}

Remember:

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

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

Negative numbers have squares too.

You may remember from Section 7.5.1: Actions that the opposite of squaring is taking the positive or negative square root, i.e. \begin{align*} x^2&= 4\\ x &= \pm\sqrt{4}\\ &= \pm 2 \end{align*} where $\pm$ means ‘plus or minus’, i.e. $x$ is plus $2$ or minus $2$ ($x=-2,2$). This works because both $-2$ and $2$ give $4$ when squared.

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

The first step is to remove the square from the lhs by taking the plus-or-minus square root of both sides: \begin{align*} x-3 &= \pm 5 \end{align*} If you like, you can split it into cases like so: \begin{gather*} \begin{aligned}[t] x-3 &= 5\\ x &= 8 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] x-3 &= -5\\ x &= -2 \end{aligned}\\ \therefore x = -2,8 \end{gather*} Or you can stick with the $\pm$ notation and write it out like this: \begin{align*} x-3 &= \pm 5\\ x &= \pm 5 + 3\\ &= -2,8 \end{align*} where $\pm 5 + 3$ means ‘$+5 + 3$ or $-5 + 3$’ which comes to $8$ or $-2$. This way is the most common notation, but some people find $\pm$ confusing (and the questions will get more complicated than this), so you can go back to writing it out in separate cases if you want.

You can also keep a comma on the rhs and whenever you do something to both sides of the equation, apply it to each possibility on the rhs like so: \begin{align*} x-3 &= -5,5\\ x &= -5+3,5+3\\ &= -2,8 \end{align*}

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

Using the $\pm$ notation: \begin{align*} (2t+3)^2 &= 49\\ 2t+3 &= \pm7\\ 2t &= \pm7 - 3\\ t &= \frac{\pm7-3}{2}\\ &= -5,2 \end{align*}

Or using cases: \begin{gather*} (2t+3)^2 = 49\\ \begin{aligned}[t] 2t+3 &= 7\\ 2t &= 4\\ t &= 2 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] 2t+3 &= -7\\ 2t &= -10\\ t &= -5 \end{aligned}\\ \therefore t=-5,2 \end{gather*}

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

If $x$ is positive or zero then $\abs{x}=x$, and if $x$ is negative then $\abs{x}=-x$, so we have two cases: \[x=5\quad\text{or}\quad -x=5.\] In other words, we replace $\abs{x}$ with $x$ or $-x$ in each case. We get two solutions: $x=5$ or $x=-5$.

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

You can set this out in separate cases according to the sign of $y+3$: \begin{gather*} \begin{aligned}[t] y+3 &\ge 0\\ y+3 &= 1\\ y &= -2 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] y+3 &\lt 0\\ -(y+3) &= 1\\ y &= -4 \end{aligned}\\ \therefore y = -4,-2. \end{gather*} Make sure you check these solutions by substituting them into the original equation.

It’s also possible to write it out like this: \begin{align*} y+3 &= \pm 1\\ y &= \pm 1-3\\ &= -4,-2 \end{align*}

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

Remember to check your solutions by substituting them into the original equation.

We can split this up into two cases depending on the sign of $x-3$: \begin{gather*} \begin{aligned}[t] x-3 &\ge 0\\ x-3 &= 2x\\ x &= -3 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] x-3 &\lt 0\\ x-3 &= -2x\\ 3x &= 3\\ x &= 1. \end{aligned} \end{gather*} But $x=-3$ came from the case where $x-3$ had to be non-negative, which is a contradiction. If we substitute $x=-3$ into the original equation, we find that the lhs is $6$ while the rhs is $-6$, so this one is an extraneous solution.

Therefore $x=1$ is the only solution.

Instead of substituting into the original equation to check for extraneous solutions, you can sub them into the inequality. For the $x=-3$ solution, $x-3 = -6\lt 0$ and this was the $x-3\ge 0$ case, so this is a contradiction. For the $x=1$ solution, $x-3 = -2 \lt 0$ and this was the $x-3\lt 0$ case, so this one is fine.

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

Since $t-2$ and $3t+4$ can each be positive or negative, it seems like there are four cases we must consider: \begin{align*} t-2 &= 3t+4\\ -(t-2) &= -(3t+4)\\ -(t-2) &= 3t+4\\ t-2 &= -(3t+4). \end{align*} But you might notice that the first and second of those equations are actually equivalent (if we multiply both sides by $-1$), and likewise with the third and fourth equation. So really there are only two cases: \begin{gather*} \begin{aligned}[t] t-2 &= 3t+4\\ -6 &= 2t\\ t &= -3 \end{aligned} \qquad\text{or}\qquad \begin{aligned}[t] t-2 &= -(3t+4)\\ &= -3t-4\\ 4t &= -2\\ t &= -\frac{1}{2} \end{aligned} \end{gather*} Therefore $t=-3,-\frac{1}{2}$.

Remember:

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