I’m cramming for a test and quadratics feel like a bowl of algebra spaghetti that keeps slipping off my fork. I know there are three big moves-factoring, completing the square, and the quadratic formula-but in my head they blend into one big, swooshy parabola smoothie.
For example, I tried to solve 3x^2 – 7x – 2 = 0. First completely-wrong attempt: I tried to cancel an x from everything (because… chaos?), so I wrote 3x^2 – 7x – 2 = 0 -> 3x – 7 – 2/x = 0, then turned it into 3x – 7 = 2/x, and then I just decided x = 2/(7 – 3) = 1. That felt very decisive and very incorrect.
Then I tried factoring and guessed (3x – 1)(x – 2) = 0 because 3x·x = 3x^2 and (-1)(-2) = 2, which is, um, not -2. I keep trying random factor pairs like I’m cracking a safe, but the safe keeps laughing at me.
Finally I went for the formula, but I keep forgetting if it’s divided by 2a or just 2. I used /2 and got some numbers that didn’t work when I plugged them back in. My brain insists the “2” looks prettier than “2a” under test pressure.
What I’m actually trying to figure out is: how do you quickly decide which method to use under a timer? Is there a fast way to tell if it’s factorable without going down a rabbit hole of guess-and-check? And when completing the square, how do you keep the signs from turning into gremlins? Any help appreciated!
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3 Responses
A fast triage that works under time: put the quadratic in ax^2 + bx + c = 0 and look at the discriminant Δ = b^2 − 4ac. If Δ is a perfect square, the roots are rational, so factoring (or the formula) will be clean; if not, don’t spend time guessing factors-use the quadratic formula. For 3x^2 − 7x − 2 = 0, Δ = 49 − 4·3·(−2) = 73, not a square, so it won’t factor over the rationals; the quadratic formula gives x = (7 ± √73)/6. A quick factorability check before that is the rational root test: try x = ±(factors of c)/(factors of a), here ±1, ±2, ±1/3, ±2/3; none give zero, so no integer factoring. If you complete the square, keep signs tame by first dividing by a, moving the constant, and adding (b/2a)^2 to both sides; here x^2 − (7/3)x = 2/3, add (7/6)^2 to get (x − 7/6)^2 = 73/36, so x = (7 ± √73)/6. To remember the denominator is 2a in the formula, note the same a appears in both −4ac and 2a. A clear walkthrough of these choices is here: https://www.khanacademy.org/math/algebra/quadratics/quad-formula-article/a/the-quadratic-formula.
Totally feel you on the “spaghetti.” My quick timer-friendly rule of thumb is: try factoring only if it looks “factorable,” and decide that fast with the ac test. For ax^2+bx+c, look for two integers that multiply to ac and add to b. If they exist, factor; if not, don’t burn time-go straight to the quadratic formula. A second quick check is the discriminant D = b^2 − 4ac: if D is a perfect square, the roots are rational, so factoring (or at least nice numbers) is likely; if not, formula time. Completing the square is great when a = 1 and b is even (because (b/2)^2 is clean), or when you want the vertex form; to keep signs tame, I like to first divide by a so the x^2 coefficient is 1, move the constant to the other side, then add the same square to both sides.
Let me try your example and think out loud: 3x^2 − 7x − 2 = 0. ac = 3·(−2) = −6; I need two integers multiplying to −6 and adding to −7… hmm: (−6,1) sums to −5, (−3,2) sums to −1, so no luck-so I wouldn’t keep guessing factors. Discriminant check: D = (−7)^2 − 4·3·(−2) = 49 + 24 = 73, not a perfect square-yep, that nudges me to the formula. So x = [−(−7) ± √73]/(2·3) = (7 ± √73)/6. If I complete the square instead (just to practice the sign-gremlin thing), I’d divide by 3: x^2 − (7/3)x − 2/3 = 0, move the constant: x^2 − (7/3)x = 2/3, add (7/6)^2 = 49/36 to both sides: (x − 7/6)^2 = 2/3 + 49/36 = 73/36, so x = 7/6 ± √73/6-same result.
Sign-keeping tips that help me under pressure: say the formula out loud in your head with “all over 2a,” and write the discriminant as a separate line so the minus signs don’t sneak past you. When completing the square, always add the same square to both sides (I literally box the added number on both sides), and keep parentheses around b^2 − 4ac until you simplify. With those habits, you can quickly decide: try ac once, peek at D; if it isn’t square, don’t wrestle-formula-fy.
Quick rule: check the discriminant Δ = b^2 − 4ac-if it’s a perfect square, factor; otherwise use the quadratic formula (denominator is 2a; for 3x^2−7x−2, Δ=73 so I’d formula-fy), and complete the square mainly after making a=1 by dividing through. Rough (maybe too rough) heuristic: if b is even or ac has small factor pairs matching b, it probably factors, but I might be over-trusting that.