Surds and extraneous roots

Calculate with exact surds, track conjugate and cancellation restrictions, and check candidates after each squaring step.

Reading settings Keep track of this lesson

Find this activity in your course

On this page

Extract square factors, not added terms

square root of 72 = square root of (36 × 2) = 6 square root of 2 and square root of 8 = 2 square root of 2, so their sum is 8√2. The square factor comes out of the root; the remaining √2 is the common exact quantity whose coefficients can be added.

Simplify before deciding which terms are alike. For example, 3 square root of 18 -2 square root of 8 = 9 square root of 2 -4 square root of 2 = 5 square root of 2. But √2 + √3 cannot be combined by adding the radicands. As a check, square root of (9 + 16) = 5, while square root of 9 + square root of 16 = 7. A square root does not distribute over a sum or difference.

For non-negative a and b, square root of ab = square root of a square root of b. Also square root of x² = |x|, not always x. At x = −3, the non-negative square root is 3.

Distribute multiplication over every term. For example,

(2 square root of 3 - square root of 2)(square root of 3 + square root of 2) = 6 + 2 square root of 6 - square root of 6 -2 = 4 + square root of 6.

Similarly, (square root of a + square root of b)² = a + b + 2 square root of ab for a,b ≥ 0. The cross term is why squaring a sum of roots does not just remove the two root signs separately.

Check your understanding 1 A square root must keep its non-negative value

For x < 0, simplify square root of 18x²/square root of 2. A proposal gives 3x. Explain whether that can have the correct sign and give an exact simplified expression.

Your extracted factors, sign condition and simplified value, or work on paper

Show answer

The numerator is 3 square root of 2 |x|, so the quotient is 3|x|. Since x < 0, this is −3x, which is positive. The proposal 3x would be negative, despite dividing two positive square roots. At x = −1, for example, the original quotient is 3 rather than −3.

Rationalise with a conjugate

1/square root of 3 = square root of 3/3, because multiplying numerator and denominator by √3 makes the denominator 3. For a two-term denominator, use its conjugate, formed by reversing the sign between the terms. For 1/(square root of 5 -2), multiply top and bottom by square root of 5 + 2:

1/(square root of 5 -2) = (square root of 5 + 2)/(5-4) = square root of 5 + 2.

The denominator becomes (√5 − 2)(√5 + 2) = 5 − 4, a difference of squares. Multiplying only the denominator would change the value rather than multiply the fraction by 1.

Keep every numerator term too. For example,

(square root of 3 + 1)/(square root of 3 -1) = ((square root of 3 + 1)²)/(3-1) = (4 + 2 square root of 3)/2 = 2 + square root of 3.

Replacing the squared numerator by 3 + 1 would omit its cross term and give a different number.

A conjugate can be zero at an allowed input

For a variable expression, check that every radicand is non-negative and every divisor used is non-zero. The expression 1/(square root of x + 2) is defined for all x ≥ 0, since its denominator is at least 2. Multiplication by its conjugate gives

1/(square root of x + 2) = (square root of x -2)/(x-4), x ≥ 0, x ≠ 4.

The extra condition x ≠ 4 comes from the conjugate √x − 2 being zero there. At x = 4 the original expression still has value 1/4, whereas the transformed fraction is undefined. Do not remove that valid input from the original expression merely because a chosen transformation fails there. State the transformed formula’s conditions and handle the original value separately if needed.

Squaring preserves candidates but can add false ones

Solve square root of (x + 1) = x-1. The radicand requires x ≥ −1, and the non-negative left side further requires x − 1 ≥ 0, so any solution must have x ≥ 1. Squaring gives x + 1 = (x − 1)², hence x(x − 3) = 0.

The algebraic candidates are 0 and 3, but 0 violates the sign condition and the original equation: 1 ≠ −1. At x = 3 both original sides equal 2. Thus the only solution is x = 3.

Squaring makes opposite signs equal, so it is not reversible without the relevant sign condition. Substitute every candidate in the original equation even after a neat factorisation.

Check your understanding 2 State the domain before squaring

Solve square root of (2x + 3) = x. State both the radicand and sign conditions, then check every algebraic candidate in the original equation.

Your input and sign conditions, squared equation, candidates and original checks, or work on paper

Show answer

The radicand requires x ≥ −3/2, while the right side must be non-negative, giving the stronger condition x ≥ 0. Squaring gives x²-2x-3 = 0, so x = 3 or −1. At 3, the original sides are both 3. At −1, they are 1 and −1, so that candidate is extraneous. Only x = 3 solves the original equation.

Isolate the remaining root before squaring again

For square root of (x + 5) + square root of (x-1) = 4, real inputs require x ≥ 1. Squaring the complete sum gives

2x + 4 + 2 square root of ((x + 5)(x-1)) = 16.

The cross term remains. Isolate it to obtain square root of ((x + 5)(x-1)) = 6-x. Its non-negative left side now requires x ≤ 6. A second squaring gives

x² + 4x-5 = (6-x)² = x²-12x + 36,

so 16x = 41 and x = 41/16. This candidate lies between 1 and 6. In the original equation its roots are 11/4 and 5/4, whose sum is exactly 4. Thus x = 41/16 is the solution. A sign check is needed at each isolated-root step, followed by substitution into the original equation.

Keep an exact answer when requested

square root of 12/square root of 3 = 2 exactly. By contrast, replacing both roots with rounded decimals can give an answer just above or below 2. Simplify symbolically first when the task requests an exact value.

Problem

Let R(x) = (square root of (x + 1) -2)/(x-3). State its original real-input conditions and find a simpler equivalent expression where the original is defined. Then solve R(x) = 1/5 and R(x) = 1/4, checking every accepted answer against the original quotient.

Write your answer, then compare it with the solution

Your original conditions, simplification, two solution sets and checks, or work on paper

Hints

Hint 1

Keep the original expression beside your algebra

Look for a difference of squares connecting the numerator’s root expression and the denominator. Decide which inputs any cancellation would use, and keep the original excluded values when solving the two equations.

Show full solution

State the original conditions and simplify
Method
Require x ≥ −1 and x ≠ 3.
Reason
The square root must be real and the original denominator cannot be zero.
Working
Answer
x-3 = (square root of (x + 1) -2)(square root of (x + 1) + 2), so R(x) = 1/(square root of (x + 1) + 2) on the original inputs. Cancelling the first factor is valid there because it vanishes only at the already excluded input 3. The simpler formula is also defined at 3, but that does not give R a new value.
Solve the first target value
Method
Set the simpler expression equal to 1/5.
Reason
It is equivalent on the original allowed inputs and has a positive denominator.
Working
Answer
square root of (x + 1) + 2 = 5 gives x = 8. It is allowed, and the original quotient is (3 − 2)/(8 − 3) = 1/5. Thus the first equation has solution x = 8.
Check what happens at the second target
Method
Set the simpler expression equal to 1/4, then return to the original conditions.
Reason
An algebraic candidate at a removed input is not a solution of the original quotient.
Working
Answer
square root of (x + 1) + 2 = 4 gives the sole candidate x = 3. The original quotient is undefined there, so R(x) = 1/4 has no solution. A value of the simplified formula at a cancelled input cannot fill the original hole.

Quick check

Compare the real-input condition, original denominator, cancelled factor and both substitutions separately. Squaring and cancellation each need their own conditions; a neat simplified expression does not erase them.

Keep track of this lesson

Mark the lesson done when you have worked through it.

Lesson progress is optional and kept on this device.

Your lesson mark records your own study. Checks are kept separately in My Learning.

Surds and extraneous roots

Read offline with Premium

Reading settings