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Wize High School Algebra I Textbook (Common Core) > Absolute Value Functions

Solving Absolute Value Equations

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Solving Absolute Value Equations

We can solve equations involving absolute values either graphically or algebraically.

We'll look at solving equations of the form ∣f(x)∣=g(x)\colorOne{|f(x)| = g(x)}.

Wize Tip
  • If the equation is not already in the form ∣f(x)∣=g(x)|f(x)| = g(x), rearrange it first!
  • Sometimes, equations don't have any solutions at all.

Solving Graphically: ∣f(x)∣=g(x)\colorOne{|f(x)| = g(x)}

Just like solving any equation graphically, we can graph the functions on both sides of the equation and look for points of intersection.

Example

Solve ∣2x+1∣−5=0|2x+1|-5=0 using a graph.

We want to get the absolute value part alone on the left, so we should start by rearranging.

Let's add 5 to both sides to get: ∣2x+10∣⏟∣f(x)∣=5⏟g(x)\underbrace{|2x+10|}_{\colorOne{\small |f(x)|}} = \underbrace{5}_{\colorFour{\small g(x)}}

Below we've graphed y=∣2x+1∣\colorOne{y=|2x+1|} in blue, and y=5\colorFour{y=5} in orange.


There are two points of intersection (POIs):

  • When x=−3x=-3, both functions produce y=5y=5, so the POI is (−3,5)(-3,5)
  • When x=2x=2, both functions produce y=5y=5, so the POI is (2,5)(2,5)
As long as you are able to graph both sides of the equation, you can get an idea of what the solutions are by looking for POIs!










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Solving Algebraically: ∣f(x)∣=g(x)\colorOne{|f(x)| = g(x)}

To solve without graphing, we use the piecewise definition of the absolute value:
∣f(x)∣={f(x),iff(x)≥0−f(x),iff(x)<0\boxed{ |f(x)| = \left\{ \begin{array}{rcl} f(x), & \text{if}& f(x)\ge0\\[0.5em] -f(x), & \text{if} & f(x)<0 \end{array} \right. }
This way, we just have to check both cases:
  1. where f(x)=g(x)f(x)=g(x) (in the case that f(x)≥0f(x) \ge 0)
  2. where −f(x)=g(x)-f(x)=g(x) (in the case that f(x)<0f(x)<0)
Example

Solve the same problem as before algebraically: ∣2x+1∣−5=0|2x+1|-5=0

Once again, we rewrite this as ∣2x+1∣=5|2x+1|=5.
According to the piecewise definition:
∣2x+1∣={2x+1,if2x+1≥0−(2x+1),if2x+1<0|2x+1|= \left\{ \begin{array}{rcl} 2x+1, & \text{if}& 2x+1\ge0\\[0.5em] -(2x+1), & \text{if} & 2x+1<0 \end{array} \right.

Simplify each of the conditions by solving for xx:

2x+1≥0    ⟹    2x≥−1    ⟹    x≥−122x+1\ge0 \ \ \implies \ \ 2x \ge -1 \ \ \implies\ \ x \ge -\frac{1}{2}
and similarly:
2x+1<0    ⟹    2x<−1    ⟹    x<−122x+1<0 \ \ \implies \ \ 2x < -1 \ \ \implies\ \ x < -\frac{1}{2}

So now we can work on each case separately by setting the parts of the piecewise function equal to the RHS, which is 5.
Case 1: 2x+1=5,  where  x≥−12\colorThree{2x+1=5,\ \ \text{where }\ x \ge -\frac{1}{2}}

Solving for xx: 2x+1=5    ⟹    2x=4    ⟹    x=22x+1=5 \ \ \implies \ \ 2x = 4 \ \ \implies\ \ \boxed{x=2}
Watch Out!
Always make sure your solution satisfies the condition!
In this case, since the condition is x≥−12x\ge -\frac{1}{2}, the solution x=2x=2 is accepted.

Case 2: −(2x+1)=5,  where  x<−12\colorThree{-(2x+1)=5,\ \ \text{where }\ x < -\frac{1}{2}}

Solving for xx: −(2x+1)=5    ⟹    −2x−1=5    ⟹    −2x=6    ⟹    x=−3-(2x+1)=5 \ \ \implies \ \ -2x-1=5 \ \ \implies \ \ -2x = 6 \ \ \implies\ \ \boxed{x=-3}
Since the condition is x<−12x < -\frac{1}{2}, the solution x=−3x=-3 is accepted.

Practice: Solving Absolute Value Equations

Select all solutions to the equation ∣x∣=4|x|=4.
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Example: Solving Absolute Value Equations

Solve ∣x2−4∣−4x=−8|x^2-4|-4x=-8.

Start by rearranging to get to the form ∣f(x)∣=g(x)|f(x)|=g(x):
∣x2−4∣−4x=−8    ⟹    ∣x2−4∣=4x−8|x^2-4|-4x=-8 \ \ \implies\ \ |x^2-4|= \colorFour{4x-8}

Writing the piecewise definition:

∣x2−4∣={x2−4,ifx2−4≥0−(x2−4),ifx2−4<0|x^2-4| = \left\{ \begin{array}{rcl} x^2-4, & \text{if}&x^2-4\ge0\\[0.5em] -(x^2-4), & \text{if} & x^2-4<0 \end{array} \right.

Now let's determine the conditions more specifically:

x2−4≥0    ⟹    x2≥4    ⟹    x≤−2  or  x≥2x^2-4\ge0 \ \ \implies \ \ x^2 \ge 4 \ \ \implies\ \ x \le -2 \ \text{ or }\ x \ge 2
and
x2−4<0    ⟹    x2<4    ⟹    −2<x<2x^2-4 < 0 \ \ \implies \ \ x^2 < 4 \ \ \implies\ \ -2 < x < 2

Now we can rewrite the piecewise definition:

∣x2−4∣={x2−4,ifx≤−2  or  x≥2−x2+4,if−2<x<2|x^2-4| = \left\{ \begin{array}{rcl} x^2-4, & \text{if}&x\le-2 \ \text{ or }\ x\ge2\\[0.5em] -x^2+4, & \text{if} & -2<x<2 \end{array} \right.

Case 1: x2−4=4x−8, where  x≤−2  or  x≥2\colorThree{x^2-4=4x-8, \ \text{where }\ x\le-2 \ \text{ or }\ x\ge2 }

x2−4=4x−8    ⟹    x2−4x+4=0    ⟹    (x−2)2=0    ⟹    x=2x^2-4=\colorFour{4x-8} \ \ \implies \ \ x^2-4x+4 = 0 \ \ \implies\ \ (x-2)^2 = 0 \ \ \implies\ \ x=2

Since the condition includes x≥2x\ge2 , our solution x=2\boxed{x=2} is valid.

Case 2: −x2+4=4x−8, where  −2<x<2\colorThree{-x^2+4=4x-8, \ \text{where }\ -2<x<2 }

−x2+4=4x−8−x2−4x+12=0−(x2+4x−12)=0−(x+6)(x−2)=0x=−6,2\begin{aligned} -x^2+4&=\colorFour{4x-8}\\[0.3em] -x^2-4x+12 &= 0\\[0.3em] -(x^2+4x-12) &= 0\\[0.3em] -(x+6)(x-2) &= 0\\[0.3em] x=-6, 2 \end{aligned}

Since the condition is −2<x<2-2<x<2, these are both extraneous (invalid) solutions!

Therefore, the only solution is x=2\boxed{x=2}.
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To solve graphically, graph y=∣x2−4∣y=|x^2-4| and y=4x−8y=4x-8.

We can see that x=2x=2 is indeed the only solution (the only point of intersection).

Practice: Solving Absolute Value Equations

Select all solutions to the equation ∣3x−6∣−x−2=0|3x-6|-x-2=0.

Practice: Solving Absolute Value Equations

Select all solutions to the equation ∣x+5∣=x2−3x|x+5|=x^2-3x.