If x² = k with k positive there are two solutions, √k and -√k. The symbol √k names one number, the non-negative one; the second solution comes from the property and not from the symbol. Keeping those two things apart is what stops half the answer going missing.
Core ideas
This method applies to any quadratic with no plain x term, and to any equation already written as a square. Both are more common than they look.
√25 is 5. It is not ±5 and it is not -5. The symbol is defined to return the non-negative root, so that taking a square root is a function with one output, and Foundations of Algebra fixed that convention. The ± in a solution is put there by you, deliberately.
x² = k is x² - k = 0, and for k ≥ 0 that is (x - √k)(x + √k) = 0 by the difference of squares. Nothing new is being assumed here; “Quadratic Equations and the Zero Product Property” is still doing the work, one layer down.
The square of any real number is 0 or positive, so x² = -9 cannot hold for a real x. It is not that the answer is hard to find — there is none among the real numbers. “Complex Numbers” builds a system in which there is one.
The material
The property, its proof, why the plus-or-minus lives outside the symbol, and the shifted form “Completing the Square” needs.
Let k be a real number. The equation x² = k has two real solutions, x = √k and x = -√k, when k > 0; the single solution x = 0 when k = 0; and no real solution when k < 0.
Suppose k > 0. Then √k is defined and positive, and k = (√k)². So x² = k is x² - (√k)² = 0, which factors as (x - √k)(x + √k) = 0 by the difference of squares. “Quadratic Equations and the Zero Product Property” splits it into x = √k or x = -√k, and those are two different numbers because √k is not 0.
If k = 0 the same factorisation reads x · x = 0, so x = 0 is the only solution — the two brackets have collapsed onto each other.
If k < 0, suppose some real x satisfied x² = k. Every real number has x² ≥ 0, so k ≥ 0, contradicting k < 0. No such x exists.
□
The ± is notation, not an operation. x = ±√k abbreviates the sentence "x is √k, or x is -√k". Writing it saves a line. Forgetting it loses a solution, and the answer that remains looks complete.
Taking the square root of both sides is often described as "square-rooting", which hides a real difficulty. √(x²) is |x|, not x: at x = -3 the left side is √9 = 3 and x itself is -3. So x² = k gives |x| = √k, and Linear Equations and Inequalities's absolute value equations then split that into the two cases x = √k and x = -√k.
Both routes — the factoring one and the absolute value one — arrive at the same pair, and both of them make the second sign a consequence of the algebra rather than a decoration added at the end.
The property is stated for x² = k, so anything standing between the square and the equals sign has to be cleared before it can be used.
Clearing the coefficient first is not the only legal order. Taking the root of both sides of 3x² = 75 gives |x|√3 = 5√3, which reduces to |x| = 5 and the same pair of answers. It is simply the order that keeps the arithmetic in plain sight, and it is the one to use.
The property does not care what is being squared. If (x - h)² = k then x - h is one of the two square roots of k, so x - h = ±√k and x = h ± √k. That is the form every completed square arrives at, which is why the whole of “Completing the Square” depends on this one.
(x - 4)² = 20. So x - 4 = ±√20, and √20 = √(4 · 5) = 2√5 by Rational and Radical Expressions. The solutions are x = 4 + 2√5 and x = 4 - 2√5, usually written together as x = 4 ± 2√5.
Check the first: (4 + 2√5 - 4)² = (2√5)² = 4 · 5 = 20. The 4s cancel before anything is squared, which is what makes checking a shifted square quick.
Two habits are worth fixing here. Simplify the radical, because 2√5 is the standard form of √20 and Rational and Radical Expressions said so. And leave it as a radical: 2√5 is a number, and replacing it with a rounded decimal replaces the answer with something that is not a solution of the equation.
x² + 9 = 0 gives x² = -9, and no real number squares to a negative. The complete and honest answer at this point in the course is "no real solution".
“Complex Numbers” introduces the number system needed to continue, and “Complex Roots of Quadratics” solves the quadratics this lesson has to set aside. The answer there is again a pair produced by this property; no unexplained symbol is needed to finish the real-number work here.
Interactive
Two roots, and the sign that is not optional
Method
The method applies when there is no loose x term, or when the square is already assembled.
4x² - 9 = 0 and 2(x + 1)² = 18 both qualify: a square, a constant, and no separate x term. x² + 6x - 7 = 0 does not, and “Completing the Square” exists to convert it into one that does.
Undo the additions and subtractions first, then the multiplication, exactly as Linear Equations and Inequalities did with linear equations. What is left must read (something)² = k, with nothing else on that side.
something = ±√k. Write the ± in the same movement as the radical, not afterwards, because afterwards is when it goes missing and because attaching it later attaches it to the wrong quantity.
Reduce √k to standard form, then finish the linear equation that is left. Substitute both answers into the original; a shifted square checks quickly, because the shift cancels before the squaring.
Each one clears everything from around the square before any root is taken.
In C the denominator was cleared by Rational and Radical Expressions's method: √5/√2 = (√5 · √2)/2 = √10/2. Leaving the answer as -3 ± √(5/2) is not wrong, but √10/2 is the standard form, and standard form is what makes two answers comparable.
A produced a symmetric pair and B and C did not. The reason is structural: A involves only x², so the equation cannot tell x from -x and its solutions must be negatives of one another. In B and C the square is shifted, and the pair is symmetric about 4 and about -3 instead. “Vertex Form and the Axis of Symmetry” gives that centre a name.
For a faded shifted-square solve, take 3(x - 2)² = 14. The supplied first line is (x - 2)² = 14/3. Take both roots, rationalize and simplify before solving for x, then check by showing the shift squares back to 14/3. Compare with x = 2 ± √42/3; a pair centred at -2 signals that the shift was moved in the wrong direction.
Given 2(x + 3)² - 5 = 0 you should isolate, take both roots in one written step, simplify the radical exactly, and check. You should also be able to say why √k on its own is not the answer: the symbol is a function, and the equation has two solutions.
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