And the roots you must throw away — an answer can look right and still be illegal.
Solve for x
x² − 9x − 3 = 5
Factor and cancel: (x + 3)(x − 3) ÷ (x − 3) = x + 3, so x + 3 = 5 and x = 2. The cancelling is legal only because x ≠ 3, and 2 respects that. D is the excluded value — it can never be a solution no matter what the algebra says.
A denominator may never be zero.
x² − 9x − 3 = 5
this must not be 0
x − 3 = 0
the value to avoid
x = 3
+ 3 both sides
x cannot be 3
You will cancel that bracket away in a moment — and then it is invisible.
Two perfect squares with a minus between them always split.
x² − 9
x · x and 3 · 3
(x − 3)(x + 3)
one minus, one plus — always in that pair
a² − b² = (a − b)(a + b)
Spotting this is the whole trick. Without factoring, the fraction cannot be simplified at all.
(x − 3)(x + 3) (x − 3) = 5
the same factor above and below cancels to 1
x + 3 = 5
x + 3 = 5 is a brand new equation. It is perfectly happy with x = 3, but the original was not. That gap is where extraneous roots come from.
x + 3 = 5
from Step 3
x = 2
− 3 both sides
You have a candidate, not an answer. Every rational equation needs one more step before you can write it down.
x = 2
2 is not 3 ✓
x = 3
banned — throw it out
The same equation with 6 on the right gives x = 3 — and that problem has no solution at all.
An extraneous root is a number the cancelled equation allows but the original forbids.
Most rational equations have several fractions. Ban the bad values, then multiply every term by the LCD so the denominators disappear. Solve xx − 1 − 2x + 1 = 4x² − 1.
x² − 1 = (x − 1)(x + 1); ban x = 1 and x = −1
Step 1, as always
x(x + 1) − 2(x − 1) = 4
× (x − 1)(x + 1), every term
x² + x − 2x + 2 = 4 → x² − x − 2 = 0
expand and tidy
(x − 2)(x + 1) = 0 → x = 2 or x = −1
two candidates
x = −1 is banned → x = 2 only
21 − 23 = 43 = 43 ✓
Clearing fractions can produce a single answer that turns out to be a banned value. Then the equation has no solution — and that is the answer to choose, not the banned number.
Solve x + 6x = 5.
ban x = 0; x² + 6 = 5x
× x, every term
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0
a quadratic
x = 2 or 3
2 + 3 = 5 ✓, 3 + 2 = 5 ✓
Different: neither candidate is banned, so both stay — testing is not the same as throwing one out.
Solve x² − 4x − 2 = x + 2.
ban x = 2; (x − 2)x + 2x − 2 = x + 2
factor the top
x + 2 = x + 2
true for every x
every real x except 2
x = 0: −4/−2 = 2 ✓
Different: the x’s vanish and leave a true statement, so there are infinitely many solutions — minus the banned one.
At x = 2 the left side is 0 ÷ 0, so 2 is not a solution even though x + 2 = x + 2 looks happy with it. The ban written at the start decides it.
A pipe fills a tank in 3 hours; a drain empties it in 5 hours. With both open, how long does the tank take to fill?
13 − 15 = 1t
tank per hour; the drain subtracts
5t − 3t = 15 → 2t = 15
× 15t, every term
t = 7.5 hours
7.53 − 7.55 = 2.5 − 1.5 = 1 tank ✓
Different: the rates are the fractions, and one of them works against the other.
One pipe takes 5 hours longer than the other to fill a tank. Together they fill it in 6 hours. How long does the faster pipe take alone?
1t + 1t + 5 = 16
faster pipe t hours; together, add the rates
6(t + 5) + 6t = t(t + 5) → t² − 7t − 30 = 0
× 6t(t + 5), every term
(t − 10)(t + 3) = 0 → t = 10 hours
a time cannot be −3; 110 + 115 = 16 ✓
Different: the unknown sits in both rates, so a quadratic appears — keep only the root that makes sense as a time.
A proportion ab = cd becomes ad = bc. Solve x + 2x − 4 = x + 5x − 2, with x ≠ 4 and x ≠ 2.
(x + 2)(x − 2) = (x + 5)(x − 4)
cross-multiply
x² − 4 = x² + x − 20
the x² terms cancel
x = 16
not banned; 1812 = 2114 = 32 ✓
Cross-multiplying 3x + 1 = 5x + 2 as if the left were one fraction is wrong. Combine the left side first, or multiply every term by the LCD as in Step 6.
Rational equations on the exam nearly always have a trap root: one candidate makes a denominator zero, and “x = a or x = b” sits among the choices next to “x = a only”. List the banned values before you start, clear the fractions, and cross out any candidate on the list.
2² − 92 − 3 = −5−1 = 5
4 − 9 = −5 and 2 − 3 = −1
Solve for x
x² − 16x − 4 = 8
Cancelling gives x + 4 = 8, so x = 4 — but x = 4 makes the original denominator zero. That value is excluded, so the equation has no solution. A is the trap: the algebra produced a number the original expression cannot accept. Always test against the excluded values.
Hint: work it exactly the same way, but look hard at the final test. Not every rational equation has an answer.
