Squaring and clearing denominators both invent solutions. Substituting back is how you catch them.
Which of the following gives all solutions to the given equation?
√3x + 4 = x
Squaring gives 3x + 4 = x², so x² − 3x − 4 = (x − 4)(x + 1) = 0 and the candidates are x = 4 and x = −1. Substitute both: x = 4 gives √16 = 4 ✓, but x = −1 gives √1 = 1, and 1 is not −1 ✗ So the answer is x = 4 only. D is the trap this whole lesson exists for — it solves the squared equation correctly and never checks. A keeps the impostor and discards the real solution. B is what you get by deciding the whole thing is impossible because one candidate failed.
Squaring only undoes a square root when the root is by itself on its side of the equals sign.
√2x + 6 − 4 = 0
the root is not alone yet
√2x + 6 = 4
+ 4 both sides
2x + 6 = 16
square both sides
x = 5
2x = 10
If you square √2x + 6 − 4 as it stands you do not get 2x + 6 − 16. A squared sum has a middle term: (A − B)² = A² − 2AB + B². Moving the 4 across first costs one line and removes the whole problem.
Whenever the other side contains an x, squaring produces an x² term. Two candidates come out; not both are guaranteed to be real solutions.
√3x + 4 = x
the radical is already alone
3x + 4 = x2
square both sides
x2 − 3x − 4 = 0
everything to one side
(x − 4)(x + 1) = 0
−4 · 1 = −4 and −4 + 1 = −3
x = 4 or x = −1
candidates, not answers
The word matters. Everything you have proved so far is that if the original equation has a solution, it is one of these two numbers. Whether either of them actually works is a separate question, and it is the one the SAT is asking.
Any x that makes a denominator zero is forbidden before you solve anything — write it down where you can see it.
3x + 2 = 5x − 2
banned: x = −2 and x = 2
3(x − 2) = 5(x + 2)
multiply by (x + 2)(x − 2)
3x − 6 = 5x + 10
distribute
x = −8
−16 = 2x
−8 is on neither banned list, so it survives. Check: 3−6 = −12 and 5−10 = −12 ✓
When one side is a sum, multiply every term by the LCD — the terms without a denominator included.
1x − 3 + 2 = xx − 3
banned: x = 3
1 + 2(x − 3) = x
× (x − 3), the lone 2 as well
2x − 5 = x
1 − 6 = −5
x = 5
not 3, so it is allowed
Squaring and multiplying by a variable are both one-way moves: they can turn a false statement into a true one, so they can hand you solutions the original never had. Those are called extraneous.
| The move | What it quietly adds | The screen |
|---|---|---|
| squaring both sides | the solutions of the equation with the other sign | a square root is never negative, so the other side must be ≥ 0 |
| multiplying out a denominator | the value that made the denominator zero | compare against the banned list |
Here is the opening problem's check, written out. Put the number into the original equation — never into the squared one, which is exactly where the impostor is telling the truth.
That third line is the whole mechanism. √3x + 4 = x and −√3x + 4 = x square to the same thing, so the squared equation carries the solutions of both. One of the two candidates belongs to the equation you were not asked about.
The SAT writes these items so that the unchecked pair of candidates is a choice. If your work produced two numbers and one of the options is simply both of them, treat that as a warning light, not a shortcut. Ten seconds of substitution decides it.
A rational equation does the same thing: two candidates come out and the banned value kills one of them.
x2x + 3 = 9x + 3
banned: x = −3
x2 = 9
× (x + 3)
x = 3 or x = −3
two candidates
x = 3
−3 is banned — it is extraneous
And when the only candidate is the banned value, the honest answer is that there is no solution. The SAT asks this as "how many solutions does the equation have".
x + 2x − 6 = 8x − 6
banned: x = 6
x + 2 = 8
× (x − 6)
x = 6
the one value that is not allowed
No solution
every candidate was extraneous
Notice how tidy that last item looks. Nothing in the algebra goes wrong — the equation simply has no legal solution, and a student who never wrote down "banned: x = 6" answers 6 with complete confidence.
With a root on each side, isolate one, square, and you will be left with a single root to isolate and square again.
√x + 7 = 1 + √x
one root on each side
x + 7 = 1 + 2√x + x
(1 + √x)² has a middle term
6 = 2√x
the x cancels
√x = 3, so x = 9
square again
Desmos settles the opening problem in one line, because a graph cannot show an extraneous root. Plot y = √3x + 4 and y = x and count the crossings.
At x = −1 the curve is at 1 and the line is at −1, so the two graphs are nowhere near each other — the dashed segment is the gap. The square-root curve only ever returns the non-negative root, which is precisely the information squaring threw away.
Graphing both sides and counting intersections answers "how many solutions" directly, and it can never report an extraneous one. Type a rational equation in the same way: y = x + 2x − 6 and y = 8x − 6 run parallel and never meet. The one thing a graph will not give you is an exact value with a radical in it, so keep the algebra for those.
Every candidate in this lesson, tested against the equation it came from.
Which of the following gives all solutions to the given equation?
x2 − 2xx − 3 = 3x − 3
The denominator bans x = 3. Multiplying through by x − 3 gives x² − 2x = 3, so x² − 2x − 3 = (x − 3)(x + 1) = 0 and the candidates are 3 and −1. The candidate 3 is the banned value, so it is extraneous; x = −1 survives, and it checks: 1 + 2−4 = −34 and 3−4 = −34 ✓ C is the unchecked pair. B picks the candidate that cancels the denominator — the one value the equation cannot take. D assumes that because one candidate died both did.
Hint: write "banned: x = 3" before you do anything else. Then clear the denominator, solve the quadratic, and compare each candidate with that note. One of the two is going to look very familiar.
