Chapter 1 · Algebraic Proofs · Concept 4
1-9 Identities
A heart monitor tracks a wave that repeats. Working out that wave in full, thousands of times a second, would flatten the battery — so engineers replace the big formula with a shorter one that gives exactly the same answer every time. That only works because the two really are the same, and showing that two different-looking expressions are the same is what this concept is about.
Guiding question
How do you show that two expressions that look different are equal for every value of x?
Learning outcome. Recognises an algebraic identity and distinguishes it from an equation.
Both of these have an equals sign. Only one of them is true for every x.
An equation
2x + 6 = 10
8 ≠ 10
breaks here
Only true at one value of x. Finding that value is what solving means.
An identity
2(x + 3) = 2x + 6
8 = 8
holds here
True at every value of x. There is nothing to solve — the two sides are the same thing written twice.
Drag the slider all the way across. The right-hand panel never breaks. That is what the word identity means.
What an identity is, and how to use it
An equality that holds true for any value put in for x is called an identity in x.
ax2 + bx + c = a′x2 + b′x + c′ is an identity in x
⟺ a = a′, b = b′, c = c′
The numbers in front of terms of the same degree are equal on both sides.
So when a question gives you an identity, put the expression in descending order of powers of x, then compare the numbers in front of each term.
For identities with fractions in them, clear the denominators first.
An identity is true for every x — not just for one or two. Two polynomials that agree at every value have to be the same polynomial, and two polynomials are the same only when every matching coefficient is the same. That is the whole reason you are allowed to split one equation into two or three little ones.
And that is what makes it easy. Instead of one hard problem you get one small problem per power of x.
x² + 6 = a(x + 1)² + b(x + 1) + c
There is no x term on the left, so its coefficient is 0 — not "nothing".
2. Find a and b so that (3x + 1)/((x+2)(x−3)) = a/(x+2) + b/(x−3) is an identity in x.
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Get rid of the fractions first.
multiply both sides by (x+2)(x−3)You cannot compare coefficients while there are fractions. This is always the first move when you see one.
Work with a partner. Do these together, then talk about the answers. Decide first whether you need to clear any fractions.
Find the values of the constants so that these are identities in x.
A toxicologist is watching how fast a compound clears from a water system. The clearance rate after t hours is
f(t) = (22t + 15) / (8t² + 10t + 3)
Split it into two partial fractions a/(2t + 1) + b/(4t + 3), and find a and b.
A missing term means a coefficient of 0.
In x² + 6 there is no x term, so its coefficient is 0 and the row reads 0 = 2a + b. Leaving that row out loses the equation you needed.
Clear the fractions before comparing.
You cannot compare coefficients while there is an x on a denominator. Multiply both sides by all the bottoms first — that is always the opening move.
Put both sides in order of powers first.
Comparing "the number in front of x" only works once the x terms are collected together. Tidy, then compare.
Do not try to solve for x.
An identity is true for every x, so there is nothing to find. What you are finding is a, b and c.
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