Algebra 1
The simplest quadratics (x²=9 → two solutions)
Here's the surprise that runs through the rest of the unit, in its plainest form. The equation x² = 9 asks: what number, squared, gives 9? Try 3: yes, 3² = 9. But now try −3: (−3)² is also 9. Two different starting numbers, one ended-up answer. Squaring lost the sign: it threw away whether you began on the positive side or the negative side, so both 3 and −3 are honest answers, and you have to report both.
That one fact, squaring loses the sign, is the whole unit in a sentence. It's why a quadratic usually has two solutions, why a little ± shows up in every method ahead, and (in the last lesson) why the U-shaped graph crosses the axis twice. Keep it in mind and the rest will feel like one idea, not a pile of separate rules.
So to solve x² = 9 you take the square root of both sides, but you write the ± yourself to put back the sign that squaring erased:
$$x^2=9\;\Rightarrow\; x=\pm\sqrt9=\pm3.$$
That ± is just shorthand for "two answers at once": x = ±3 means x = 3 or x = −3 12.2.f1. This whole move has a name, the Square-Root Property, and it has two branches, depending on the sign of the number you're taking the root of.
When that number is zero or positive, you get the two roots (collapsing to a single x = 0 when the number is 0). When it's negative, there's no real solution at all, because no real number squared can land on a negative. That second branch isn't a special exception to patch over later. It's half of the rule, and you'll lean on it again before the unit is out.
The negative branch is worth seeing once now so it doesn't rattle you later. Take x² = −4. No real number squared gives −4, because every real square is zero or positive (2² = 4 and (−2)² = 4; nothing real reaches −4). So the answer is "no real solution," and that is the answer. It's a finished, correct result, not a sign you did something wrong.
Two practical notes before the examples. First, isolate the x² before you take a root. If you see 2x² = 50, divide by 2 first to get x² = 25, then root to get x = ±5. Don't root the whole thing while the 2 is still attached. Second, when the number isn't a perfect square, the answer stays in radical form, exactly as in Lesson 12.1: x² = 7 gives x = ±√7, and that's exact and done.
New words
- 12.2.d1 ±("plus or minus"): shorthand for two answers at once. x=±3 means x=3 or x=-3.
- 12.2.d2 Quadratic equation: an equation with an x² term (and no higher power). Standard form is ax²+bx+c=0 with a≠0. These are the special case where b=0.
- 12.2.d3 Square-Root Property (the rule of this lesson, worth boxing): for x²=k,
- if k≥0, then x=±√k (i.e. x=√k or x=-√k, two real solutions, collapsing to the single x=0 when k=0);
- if k<0, there is no real solution (no real number squares to a negative).
Stated once, conditional and complete: x²=k ⇒ x=±√k when k≥0; no real solution when k<0. The k<0 branch is not a side-exception to patch later. It is half of the rule, and it returns in 12.5 as the discriminant's sign.
Worked example
- 12.2.w1 x²=9 ⇒ x=±3, i.e. x=3 or x=-3. Check both: 3²=9, (-3)²=9. Both starting numbers square to 9, which is exactly why both are answers.
- 12.2.w2 x²=16 ⇒ x=±4. Same move: 16 is 4², so the two roots are 4 and −4.
- 12.2.w3 x²-7=0 ⇒ x²=7 ⇒ x=±√7 (exact, irrational, not "unfinished," per 12.1). Add 7 to both sides to isolate x², then root; 7 isn't a perfect square, so √7 stays.
- 12.2.w4 2x²=50 ⇒ x²=25 ⇒ x=±5. (Isolate x² first, dividing by 2, before rooting.) If you rooted before dividing you'd drag along an extra radical and land on ±5√2, which is wrong here.
- 12.2.w5 x²-3=0 ⇒ x²=3 ⇒ x=±√3. Isolate, then root; 3 isn't a perfect square, so leave the radical.
- 12.2.w6 The k<0 branch (no real solution): x²=-4. Here k=-4<0. By the Square-Root Property there is no real solution, because no real number squares to -4, since any real square is ≥0. (Check the intuition: 2²=4 and (-2)²=4; nothing real lands on -4.) The answer is "no real solution," full stop. Same for x²+4=0 ⇒ x²=-4: isolate first, see the negative, stop.
The substitution check above is how you'll know you're right when no one's there to tell you. If you put an answer back and the two sides don't match, you haven't failed. Your check just did its job and caught something before it counted. Go back to your first step and re-run the arithmetic slowly; a mismatch is almost always one sign or one small slip, not the whole method.
Here's a clean one to get the rhythm before the set turns mixed: solve x² = 1. The square root of 1 is 1, so x = ±1, meaning x = 1 or x = −1. Check: 1² = 1 and (−1)² = 1. Both land, both count.
One slip deserves a callout now that you've done several correctly: writing only x = 3 for x² = 9 and stopping. That's the most natural mistake here, and it comes from forgetting that squaring hid a sign. The fix is built into the method. Whenever you undo a square, write the ± first, before you fill in the number, so the second root can't slip away. (And if isolating ever leaves you with x² equal to a negative, don't force a ± onto it; the negative branch already told you the honest answer is "no real solution.")
Check yourself
- 12.2.c1 Solve x² = 49. How many answers, and why isn't it just one? (x = ±7, so two answers, x = 7 or x = −7, because both 7² and (−7)² equal 49; squaring lost the sign, so both starts are valid.)
- 12.2.c2 A friend says x² = 20 has answer x = √20. What did they leave out, and can you simplify the radical too? (They left out the negative root: it's x = ±√20. And √20 = √(4·5) = 2√5, so the full answer is x = ±2√5.)
- 12.2.c3 Solve 3x² = 27. What has to happen before you take a square root? (Divide both sides by 3 to isolate x², giving x² = 9, and only then root: x = ±3. Isolating the x² comes first.)
You can now solve x² = k by isolating the x², writing the ± to keep both signs when k is zero or positive, and recognizing "no real solution" as the correct, finished answer when k is negative.
Mixed practice feels harder, and that's the point. It's what makes this stick. Every problem has its answer at the end of the lesson; if one stalls you, flip back to the worked example it's based on.
Solve (keep both signs; leave irrationals exact):
Reveal answerHide to problem 1
x=±5Reveal answerHide to problem 2
x=±7Reveal answerHide to problem 3
x=±1Reveal answerHide to problem 4
x²=11, x=±√11Reveal answerHide to problem 5
x²=2, x=±√2Reveal answerHide to problem 6
x²=9, x=±3Reveal answerHide to problem 7
x²=9, x=±3Reveal answerHide to problem 8
x²=9, x=±3Reveal answerHide to problem 9
x²=100, x=±10Reveal answerHide to problem 10
x=±√5No real solution (the k<0 branch). State it and say why: