Algebra 1
The quadratic formula & the discriminant
Factoring is quick when it works, and completing the square always works but is fiddly to redo every time. The quadratic formula is the way out: it's completing the square done once, in general, so you never have to grind through it again. You just read the three numbers a, b, and c off the equation and drop them in. It solves every quadratic, factorable or not, with rational answers or irrational ones. That's why it's the universal tool.
For any quadratic written as ax² + bx + c = 0,
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
That ± is the same one from every lesson in this unit: it's recovering the sign that squaring threw away. To use the formula well, read a, b, and c carefully, with their signs: for 2x² + 3x − 2 = 0, a = 2, b = 3, and c = −2, where the minus belongs to the c.
Here's where the 2a on the bottom comes from, if you're curious. The general derivation first divides everything by a so the x² coefficient becomes 1, then completes the square (Lesson 12.4), and that division is exactly what leaves the 2a in the denominator. You don't have to redo it; the formula already did.
There's a bonus tucked inside the formula. The part under the root, b² − 4ac, is the discriminant, and its sign tells you how many real solutions you'll get before you finish the arithmetic.
If it's positive, the root of a positive number gives two different values, so two real solutions 12.5.f1. If it's zero, the ± adds and subtracts nothing, so the two solutions collapse into one (a repeated root). If it's negative, there's no real square root to take, so no real solution. That negative case is the same "no real solution" branch you met in Lesson 12.2, now showing up as the sign of the discriminant.
One refinement when the discriminant is positive, worth knowing because it explains a pattern you've already seen. If b² − 4ac is itself a perfect square, the root comes out whole and the two answers are rational, which means the quadratic would have factored over the integers (the Lesson 12.3 case). If it's positive but not a perfect square, the answers keep a radical and are irrational.
That's exactly why some quadratics factor cleanly and others only yield to the formula. (For instance: x² − 3x − 10 = 0 has discriminant 49 = 7², so rational roots 5 and −2; x² − 2x − 5 = 0 has discriminant 24, so irrational roots 1 ± √6.)
The working order is steady: substitute a, b, c; simplify the discriminant first; then take its root and resolve the ±. Then check a root by putting it back in.
New words
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12.5.d1 Quadratic formula: for ax²+bx+c=0 (with a≠0), $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ The ± is the same two-solutions ± as always, recovering the sign squaring destroyed.
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12.5.d2 Discriminant: the inside-the-root part, b²-4ac. For a real quadratic (a≠0, real coefficients) its sign counts the real solutions before you finish:
- b²-4ac>0: two real solutions (root of a positive → ± two values).
- b²-4ac=0: one real solution, a repeated (double) root (±0 collapses to one; the parabola's vertex sits on the axis).
- b²-4ac<0: no real solution (no real square root of a negative; the parabola misses the axis).
- One refinement when b²-4ac>0 (optional, real-only): a perfect-square discriminant means the two roots are rational, because the √ comes out whole, so the quadratic factored over the rationals (the 12.3 case). A non-perfect-square discriminant means the two roots are irrational, keeping a radical (12.5's payoff, e.g. 1±√6). This is exactly why some quadratics factor cleanly and others only yield to the formula. (Concretely: x²-3x-10=0 has discriminant 49=7², rational roots 5,-2; x²-2x-5=0 has discriminant 24, irrational roots 1±√6.)
Worked example
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12.5.w1 x²-5x+6=0: a=1,b=-5,c=6. Discriminant (-5)²-4(1)(6)=25-24=1>0 → two solutions. Note -b is +5 here because b is -5, and (-5)² is +25. $$x=\frac{-(-5)\pm\sqrt1}{2}=\frac{5\pm1}{2}=\{3,\,2\}.$$ (Matches the factoring answer from 12.3; every method agrees.)
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12.5.w2 2x²+3x-2=0: a=2,b=3,c=-2. Discriminant 3²-4(2)(-2)=9+16=25 → two. Watch the -4ac: with c negative, -4(2)(-2) adds 16. $$x=\frac{-3\pm\sqrt{25}}{4}=\frac{-3\pm5}{4}=\left\{\tfrac{2}{4}=\tfrac12,\ \tfrac{-8}{4}=-2\right\}.$$
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12.5.w3 x²-4x+4=0: a=1,b=-4,c=4. Discriminant (-4)²-4(1)(4)=16-16=0 → one solution. $$x=\frac{4\pm\sqrt0}{2}=\frac{4}{2}=2.$$ (A repeated root: the perfect square (x-2)²=0. The ± adds nothing because √0 is 0.)
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12.5.w4 x²+x+1=0: a=1,b=1,c=1. Discriminant 1²-4(1)(1)=1-4=-3<0 → no real solution (can't take a real square root of -3). We stop here, and that's the finished, correct answer.
A clean one to settle the formula before the set mixes things up: solve x² − 5x + 4 = 0. Here a = 1, b = −5, c = 4, so the discriminant is (−5)² − 4(1)(4) = 25 − 16 = 9, a perfect square, so expect two rational roots. Then x = (5 ± √9)/2 = (5 ± 3)/2, giving x = 4 or x = 1. Check: 4² − 5(4) + 4 = 0, and 1² − 5(1) + 4 = 0.
Now the slips this formula invites, named after a few clean runs. The biggest by far is −b when b is already negative: −b means the opposite of b, so if b = −5 then −b = +5, not −5.
Next, sign errors in the discriminant, especially −4ac when c is negative, which makes that piece add (as in example 2). And remember that b² is always positive even when b is negative ((−5)² = 25; this is the same care as telling −3² from (−3)²).
Another is dividing only part of the top by 2a: the entire −b ± √( ) sits over 2a, so keep the fraction bar running full width. And the negative-discriminant result, "no real solution," is a real and correct finding, not a mistake to fix; the graph in the next lesson will show the U missing the axis entirely.
If a check ever comes out wrong, that's the safety net working, not a verdict. Re-run the arithmetic from the first step. Most often it's a single sign, the −b or one piece of the discriminant, that slipped.
Check yourself
- 12.5.c1 Use the formula on x² − 2x − 5 = 0. What's the discriminant, and what does its sign tell you before you finish? (a = 1, b = −2, c = −5; discriminant (−2)² − 4(1)(−5) = 4 + 20 = 24 > 0, so two real solutions, and since 24 isn't a perfect square they'll be irrational. Finishing: x = (2 ± √24)/2 = 1 ± √6.)
- 12.5.c2 Without solving, how many real solutions does x² + x + 5 = 0 have? How can you tell? (Discriminant 1² − 4(1)(5) = 1 − 20 = −19 < 0, so no real solutions. A negative discriminant means there's no real square root to take.)
- 12.5.c3 In 2x² − 7x + 3 = 0, what are a, b, and c, signs included, and what is −b? (a = 2, b = −7, c = 3; and −b = −(−7) = +7. The sign travels with each number.)
You can now solve any quadratic with the formula by reading off a, b, c with their signs, simplifying the discriminant first, and resolving the ±. And you can read the discriminant's sign to know how many real solutions to expect, and whether they're rational or irrational.
Mixed practice feels harder on purpose, and that's what makes it last to next week. Every answer is at the end of the lesson; if one stalls you, flip back to the worked example it's based on.
Find the discriminant and state the number of real solutions:
Reveal answerHide to problem 1
9+40=49>0, two solutions (x=5,-2)Reveal answerHide to problem 2
36-36=0, one solution (x=-3)Reveal answerHide to problem 3
1-20=-19<0, no real solutionsReveal answerHide to problem 4
36-40=-4<0, no real solutionsSolve with the quadratic formula: