A reminder of the vocabulary we lean on, so nothing here is unearned:
Three pictures anchor this whole page. Keep them in your mind's eye as you work through every trap below.
Picture 1 — roots are x-intercepts. A root isn't an abstract number; it is a place where the curve meets the ground. Look at the blue parabola: the two red dots sit exactly where the height is zero.
Picture 2 — why a product being zero pins down the intercepts. Factoring writes the curve as (x−p)(x−q). The two straight lines below are the factors x−2 and x−3; each hits zero at one place, and the parabola (their product) can only be zero where one of the lines is zero. That is the Zero Product Property drawn out.
Picture 3 — completing the square centers the parabola. The yellow arrow shows the shift x→x+2ab that slides the curve so its lowest point sits on the vertical axis. Once centered, "how far to the roots" is just a symmetric ± step — that symmetry is the whole trick.
Picture 4 — the three fates set by the discriminant. The same three cases the True/False items keep mentioning, drawn side by side: the blue curve crosses the ground twice (discriminant positive, two real roots), the yellow curve just touches it once (discriminant zero, one repeated root), and the red curve misses the ground entirely (discriminant negative, no real roots).
An equation with no visible x2 term written can never be quadratic
False — x(x+3)=4x−6 hides an x2; you must expand and collect terms first. Only after standard form ax2+bx+c=0 can you check whether a=0.
Every quadratic equation has two real roots
False — it has two roots counting multiplicity and complex numbers, but the number of real roots can be 2, 1, or 0 depending on whether the parabola crosses, touches, or misses the axis (all three drawn in Picture 4, governed by the sign of the discriminant b2−4ac).
If x2=9 then x=3
False — squaring destroys sign information, so x=3orx=−3. Forgetting the ± throws away half the solutions.
(x−2)(x−3)=0 and x2−5x+6=0 have exactly the same roots
True — they are the same equation, just written factored versus expanded. Expanding one gives the other, so the roots 2 and 3 are shared (this is exactly the pair of lines in Picture 2).
Completing the square only works when factoring fails
False — completing the square works on every quadratic, including easily-factorable ones. It's a universal method, not a backup.
If (x−2)(x−3)=2 then x−2=2 or x−3=2
False — the Zero Product Property needs the product to equal zero, not 2. You must first expand, move the 2 over, and re-factor x2−5x+4=0.
Multiplying a whole quadratic equation by a nonzero constant changes its roots
False — x2−5x+6=0 and 3x2−15x+18=0 have identical roots; the constant divides out because the right side is 0.
The number k we add to complete the square is always (2b)2
True only after the x2-coefficient has been made 1. The rule "k=(2b)2" is really "k=(2coefficient of x)2when that coefficient sits in front of a bare x2." For ax2+bx+c with a=1 you must first divide the whole equation by a (or factor a out), which replaces b by ab; then the number to add is (2ab)2, not (2b)2. Using b before normalizing halves the wrong quantity.
The completing term 9 was added only to the left side. An equation is a claim that two sides are equal; adding 9 to just one side breaks that equality, so the new line is a different (false) statement. To keep the same solution set you must add 9 to both: −5+9=4.
x2−5x+6=(x−2)(x−3), so the roots are x=−2 and x=−3 — find the mistake
The signs were copied instead of solved. A root is the value that makes a factor zero, so ask "what x makes x−2=0?" — that is x=+2, not −2. The roots are +2 and +3.
2x2+8x+6=2(x+2)2 — find the mistake
(x+2)2=x2+4x+4, which forces a +4 that isn't in the original constant 3. Correct: 2[(x+2)2−1] — the leftover −1 was dropped.
3x2+6x=0⇒ divide by 3x to get x+2=0, so x=−2 — find the mistake
Dividing by 3x is only legal when 3x=0, but x=0 is one of the very roots we're hunting — so the division quietly throws away a genuine solution. Never divide an equation by an expression that can be zero; factor instead: 3x(x+2)=0 gives x=0 or x=−2.
To solve x2=5x, cancel x to get x=5 — find the mistake
Cancelling x divides by a quantity that could be zero; the Zero Product Property only works if you keep both factors, and dividing erases the x=0 factor. Move everything over instead: x2−5x=0⇒x(x−5)=0, giving x=0 and x=5.
(x+3)2=4⇒x+3=2⇒x=−1 — find the mistake
Only the positive square root was taken. A square equal to 4 comes from ±2, so also x+3=−2⇒x=−5.
For ax2+bx+c=0 we found x=a−b±b2−4ac — find the mistake
The denominator should be 2a, not a. Completing the square produces a 2a1 from halving ab, and it never cancels.
4x2−9=0⇒(2x−3)(2x+3)=0⇒x=3,x=−3 — find the mistake
The factors give 2x−3=0⇒x=23, not 3. Solving each factor requires dividing by the leading coefficient inside it.
Why does setting each factor to zero give all the roots and never extra false ones?
Because the Zero Product Property is an if-and-only-if: a product is zero exactly when a factor is zero. Picture 2 makes this literal — the parabola touches the ground only above the spots where one of the straight-line factors touches the ground, and nowhere else.
Why must a=0 in the definition of a quadratic?
If a=0 the x2 term vanishes and the equation collapses to the linear bx+c=0 — a straight line, one root, handled by linear methods, not this topic.
Why do we add and subtract the same term when completing the square instead of just adding it?
Adding then subtracting the same number changes nothing about the value — it's a disguised zero. It merely regroups the expression so a perfect square becomes visible.
Why does completing the square, done in full generality, produce the quadratic formula?
Because the formula is the completed-square result with the letters a,b,c left symbolic. Running the procedure once on the general equation yields x=2a−b±b2−4ac every time, and the discriminant b2−4ac is precisely what sits under that root — see the discriminant note.
Why is (x+2ab)2 the natural square to aim for and not some other?
Because a perfect square (x+k)2 has middle term 2kx. Matching 2k=ab forces k=2ab — it's the only k that reproduces the linear term already present.
Why does "completing the square" deserve the word centering?
Picture 3 shows it: the substitution x→x+2ab slides the parabola sideways until its vertex sits on the vertical axis. Once centered, the two roots are mirror images at ± the same distance from the axis, which is why the final answer is a symmetric −2ab±(something).
Why does the AC method work for non-monic quadratics?
Because splitting the middle term into two pieces that multiply to a⋅c and add to b recreates the cross-terms of a factored product (mx+p)(nx+q); grouping then pulls out the shared binomial — the hidden factorisation. The "AC" in the name is literally the product a⋅c you start from (see the AC-method definition box above). More in grouping techniques.
Why is the completed-square form a(x−h)2=k useful beyond solving?
The h locates the parabola's vertex (axis of symmetry) directly — it's exactly the shift drawn in Picture 3 — connecting to vertex form and showing solving and graphing are the same act seen two ways.
Why can factoring sometimes fail even though the equation has real roots?
Factoring "by inspection" needs nice (rational) roots. Irrational roots like 24±10 exist and are real but won't come from integer factor pairs — completing the square or the formula is required.
A single root x=0, called a double root — the parabola just touches the axis at its vertex rather than crossing it (the yellow "touch" curve in Picture 4). Two equal roots, not one missing.
Can a quadratic have exactly one root?
Only in the "repeated" sense: x2−6x+9=(x−3)2=0 gives x=3 twice. There's one distinct value but multiplicity two, exactly when the discriminant b2−4ac is zero.
What happens when you complete the square and the right side turns out negative, e.g. (x−2)2=−3?
No real solution exists, because a real square is never negative. The parabola never reaches the axis (the red "miss" curve in Picture 4); the roots are complex — this is the negative-discriminant case, previewed in the discriminant note.
How do you handle x2+bx=0 with c=0?
Factor out x immediately: x(x+b)=0, giving x=0 or x=−b. Never divide by x — that would lose the guaranteed root at zero.
How do you handle ax2+c=0 with b=0 (no middle term)?
Skip factoring entirely: isolate x2=−ac and square-root with ±. Real roots exist only when −ac≥0.
If both roots are the same number, what does the factored form look like?
A single squared factor: a(x−p)2=0. Seeing a perfect square in factored form is the signature of a double root.
Is 0.5x2+x+0.5=0 a valid quadratic even though a is a fraction?
Yes — a only needs to be nonzero, not an integer. Fractional or irrational coefficients are fine; you may clear fractions by multiplying through (roots unchanged) for convenience.
Recall One-line self-test
Cover everything above. Can you state the one property that makes factoring work, the one condition that makes an equation quadratic, the one value you add to complete the square (and the normalization it requires), and what the sign of the discriminant tells you?
Answer ::: Zero Product Property; a=0; add (2b)2only after dividing by a so the x2-coefficient is 1 (so for general ax2+bx+c the added term is (2ab)2); and the sign of b2−4ac gives two / one / zero real roots for positive / zero / negative.
