Math 20-1 · Absolute value and radicals
Absolute value and radicals
Two topics that look like notation and are really about one idea: some operations throw away information, and when you undo them you have to put it back by hand.
- 1. The words, first
- 2. Absolute value as distance
- 3. Simplifying and operating on radicals
- 4. Radical equations and extraneous roots
- 5. What costs marks
The words, first
The idea: Absolute value is a distance. A radical is the inverse of a power. Both are one-way streets, and that is where the traps come from.
| Word | What it means |
|---|---|
| Absolute value | How far a number is from zero, ignoring direction. |−6| = 6 and |6| = 6. It is not “make it positive” when there is a variable inside — it is “whichever of x and −x is not negative”. |
| Radical | A root sign. √x is the principal square root, which means the non-negative one, so √9 is 3 and not ±3. |
| Radicand | Whatever sits under the root sign. In √(2x + 3) the radicand is 2x + 3. |
| Index | The small number in the crook of the sign saying which root: ³√8 = 2. No number written means an index of 2. |
| Entire radical | A radical with everything inside: √50. |
| Mixed radical | A number multiplied by a radical: 5√2. The same value, written so the radicand is as small as possible. |
| Like radicals | Radicals with the same index and the same radicand. Only like radicals can be added, exactly as only like terms can. |
| Conjugate | The same binomial with the middle sign flipped: the conjugate of 3 − √5 is 3 + √5. Multiplying the two kills the radical. |
| Restriction | A limit on what x is allowed to be. An even root needs a radicand of zero or more, so √(x − 5) restricts x to x ≥ 5. |
| Extraneous root | A value that satisfies the squared equation but not the original one. It is not a mistake in your algebra; it is a side effect of squaring. |
Absolute value as distance
The idea: |a − b| is the distance between a and b on the number line. Reading it that way turns most absolute value questions into a sentence about distance.
|x| is defined piecewise, and writing it out removes the mystery:
|x| = x when x ≥ 0 · |x| = −x when x < 0
That second line looks wrong the first time. It is not: if x = −6, then −x = 6, which is exactly the answer you want. The minus sign is what makes it positive.
Solving |2x − 5| = 9. Something has absolute value 9, so that something is either 9 or −9. Two equations, two answers:
2x − 5 = 9 → x = 7 · 2x − 5 = −9 → x = −2
Both check. Solving only the first is the most common way to lose half the marks on this kind of question.
When the other side has a variable, check. For |x − 3| = 2x the two cases give x = −3 and x = 1. But at x = −3 the right side is −6, and an absolute value can never be negative, so that one is rejected. Checking is not a formality here.
Reading it as distance. |x − 4| < 3 says x is less than three units from 4, so 1 < x < 7. And |x − 4| > 3 says more than three units away, so x < 1 or x > 7 — two separate pieces. Less-than gives one interval; greater-than gives two.
Simplifying and operating on radicals
The idea: Pull out perfect squares, then treat √2 as though it were a variable. Everything about adding, subtracting and collecting follows from that.
Entire to mixed. Find the largest perfect square that divides the radicand: √50 = √(25 · 2) = 5√2. Going back the other way, 5√2 = √(25 · 2) = √50 — square the outside number and take it in.
Adding. √18 + √8 is not √26. Simplify first: 3√2 + 2√2 = 5√2. If the radicands do not match after simplifying, the sum cannot be collected at all.
Multiplying. Outside times outside, inside times inside: (2√3)(5√6) = 10√18 = 10 · 3√2 = 30√2. Always simplify the result — 10√18 is the same number but not the final form.
Rationalizing a single term. Multiply top and bottom by the radical: 6/√3 = 6√3/3 = 2√3. You are multiplying by √3/√3, which is 1, so nothing about the value changes.
Rationalizing a binomial. Use the conjugate, because (a − b)(a + b) = a² − b² kills the root:
4/(3 − √5) × (3 + √5)/(3 + √5) = 4(3 + √5)/(9 − 5) = 4(3 + √5)/4 = 3 + √5
Radical equations and extraneous roots
The idea: Squaring both sides is legal and lossy: it erases the sign, so the squared equation can have solutions the original never had. You get them back out by checking.
Method. Isolate the radical, square both sides, solve, then substitute every answer back into the original equation.
Worked example: √(2x + 3) = x.
- Square: 2x + 3 = x².
- Rearrange: x² − 2x − 3 = 0.
- Factor: (x − 3)(x + 1) = 0, so x = 3 or x = −1.
- Check x = 3: √9 = 3 ✓. Check x = −1: √1 = 1, but the right side is −1. ✗
So the solution is x = 3 only. Note why −1 appeared: squaring turned “equals x” into “equals x or −x”, and −1 is a solution to the second one.
Restrictions are a free check. For √(2x + 3) = x the radicand needs 2x + 3 ≥ 0, so x ≥ −1.5, and the right-hand side must be non-negative, so x ≥ 0. Writing that down first tells you −1 cannot possibly work before you do any algebra.
What costs marks
The idea: Almost every lost mark in this unit is one of four things.
- Solving only the positive case of an absolute value equation. There are two cases until you prove otherwise.
- Adding radicands. √18 + √8 ≠ √26. Simplify, then collect like radicals.
- Not checking radical solutions. Squaring manufactures roots. An unchecked answer is an unfinished question.
- Dropping restrictions. If a question asks for the restrictions, they are worth marks on their own, and they often rule out an answer for free.