Radical Functions & Equations — Print

Squaring both sides is legal, and it can still hand you an answer that is wrong.

What the practice questions cover

These questions use 5 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Solve √x = -3.

Square both sides. Squaring is something you are allowed to do to an equation, so nothing improper has happened.

every step was legal, and the answer is wrong the algebra √x = −3 square both sides x = 9 the check √9 = 3 but we needed −3 squaring forgets the sign it started with the last line of a radical equation is the check

You get x = 9. Now put it back: √9 = 3, and 3 is not -3.

Every step was legal, and the answer is still wrong. Something about squaring loses information.

Squaring forgets the sign it started with — 3 and -3 both square to 9. So a radical equation is not finished until every answer has been put back and checked.

Simplifying a square root by pulling out a perfect square pull out the largest perfect square sqrt 72 sqrt(36 . 2) sqrt36 . sqrt2 6 sqrt2 start square factor split the root done 36 is the biggest perfect square dividing 72; check it: 6^2 . 2 = 36 . 2 = 72 using 4 instead gives 2 sqrt18, which is right but not yet finished

The shape of a root graph

y = √x starts at the origin and climbs, but more and more slowly — it is half a parabola on its side.

It exists only for x ≥ 0, because a square root of a negative is not a real number.

y = √[3]x has no such limit. It accepts negatives and runs through all four quadrants.

Where a root graph starts

For y = √x - 4, the inside must be ≥ 0, so the graph begins at x = 4. Set the inside to zero and you have the starting point.

Isolate the root first

Get the radical alone on one side before squaring. Squaring a sum squares the cross terms too, and creates work you did not need.

√x+1 + 2 = 6 should become √x+1 = 4 before anything else happens.

Then square once, cleanly, and solve what is left.

Square the sides, not the terms

(√x + 2)2 is x + 4√x + 4, not x + 4. Squaring does not distribute over addition, which is exactly why you isolate first.

Always check

Squaring can create solutions that were never there, so the last step of every radical equation is to put the answers back.

This is not neatness — it is part of the method. An unchecked answer is an unfinished problem.

If a value fails, say so and discard it. Sometimes every candidate fails, and the honest answer is "no solution".

A one-line filter

A square root can never equal a negative number. If the isolated equation reads √… = -5, you can stop right there.

Rational exponents say the same thing

x1/2 means √x, and x2/3 means √[3]x2. The denominator is the root, the numerator is the power.

Writing roots as exponents lets you use the exponent rules instead of learning separate radical rules.

√x · √[3]x is awkward; x1/2 · x1/3 = x5/6 is not.

Bottom is the root

82/3: cube root first to get 2, then square to get 4. Taking the root first keeps the numbers small.

Two radicals

When two roots appear, isolate one, square, and see what survives. Usually one radical is left.

Isolate that one and square again. Two squarings are normal and nothing has gone wrong.

Checking matters even more here, because each squaring is another chance to invent a solution.

Expect it to get worse before better

After the first squaring the equation often looks messier. That is the method working, not a sign you took a wrong turn.

Worked examples

A straightforward one

Solve √2x + 3 = 5.

  1. The radical is already alone, so square both sides: 2x + 3 = 25. Squaring undoes the root.
  2. Solve: 2x = 22, so x = 11. Ordinary linear algebra from here.
  3. Check: √2(11)+3 = √25 = 5. It works. The check is part of the method, not an optional extra.

x = 11

One that lies to you

Solve √x + 6 = x.

  1. Square: x + 6 = x2, so x2 - x - 6 = 0. A quadratic appears — that is normal.
  2. Factor: (x-3)(x+2) = 0, giving x = 3 or x = -2. Two candidates, not yet two answers.
  3. Check x=3: √9 = 3, good. Check x=-2: √4 = 2 ≠ -2, so discard it. The root is positive; it can never equal -2.

x = 3 only

A rational exponent

Evaluate 163/4 without a calculator.

  1. The denominator 4 is the root: √[4]16 = 2. Take the root first to keep the numbers small.
  2. The numerator 3 is the power: 23 = 8. Power second.
  3. So 163/4 = 8. Doing it the other way needs 163 = 4096 first — same answer, much more work.

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Words to know

Radical equation
An equation with the unknown underneath a root sign.
Extraneous solution
A value the algebra produces that fails in the original equation.
Principal root
The non-negative one. √25 = 5, never -5.
Index
The little number on the root. Square root has index 2, cube root index 3.
Domain restriction
A square root needs its inside to be ≥ 0, which limits the inputs allowed.

Mistakes to watch for

Skipping the check. Solving √x+6 = x gives x = 3 and x = -2, but -2 fails because a square root is never negative. Squaring can invent solutions, so the check is part of the method.

Squaring term by term. (√x + 2)2 is x + 4√x + 4, not x + 4. Squaring does not distribute over addition, which is why the radical is isolated first.

Reading x2/3 as 23x. The fraction is an exponent: the bottom is the root and the top is the power, so 82/3 = 4, not 163.

Ignoring the domain of the graph. y = √x-4 does not exist to the left of 4, so plotting points at x = 0 or x = 2 produces a curve that is not there.

Giving up after one squaring when two radicals are present. One squaring usually leaves a single radical behind. Isolate it and square again — the mess in the middle is expected.

Answers like this have a name: extraneous solutions. They are not mistakes in your working — they are created by a legal step that is not reversible.

The square root sign means the positive root only. Without that convention √9 would be two numbers, and √x would not be a function at all.

Cube roots have no such trouble. √[3]-8 = -2 is perfectly happy, because cubing keeps the sign.

Problem 1

Solve √x=7.

Worked answer

49

x=49

Check: √49=7

Problem 2

Solve √x+5=4.

Worked answer

11

x+5=16

x=11

Problem 3

At what x does y=√x-3 begin?

Worked answer

3

x-3 ≥ 0

x ≥ 3

Problem 4

Evaluate 251/2.

Worked answer

5

√25=5

Problem 5

Evaluate 271/3.

Worked answer

3

√[3]27=3

Problem 6

Solve √2x-1=5.

Worked answer

13

2x-1=25

2x=26

x=13

Problem 7

Solve √x+3=8.

Worked answer

25

√x=5

x=25

Problem 8

Evaluate 82/3.

Worked answer

4

√[3]8=2

22=4

Problem 9

Evaluate 163/4.

Worked answer

8

√[4]16=2

23=8

Problem 10

Solve √x+6=x. Give the only valid answer.

Worked answer

3

x+6=x2

(x-3)(x+2)=0

x=-2 fails since a root is never negative

Problem 11

How many solutions does √x=-4 have?

  1. None
  2. One
  3. Two
  4. Infinitely many
Worked answer

None

A principal square root is never negative

Squaring gives x=16, but √16=4 ≠ -4

Problem 12

Solve √3x+1=x-1. Give the only valid answer.

Worked answer

5

3x+1=x2-2x+1

x2-5x=0 so x=0 or x=5

x=0 gives 1=-1, which fails

Problem 13

Solve √[3]x= -2.

Worked answer

-8

Cube both sides

x=-8

Problem 14

Write √[5]x3 using a rational exponent.

Worked answer

x^(3/5)

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Problem 15

Solve 2√x-1 = 9.

Worked answer

25

2√x=10

√x=5

x=25

Problem 16

For what values of x is √5-x defined? Give the largest allowed x.

Worked answer

5

5-x ≥ 0

x ≤ 5

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