Algebra 1 · A.REI.B.4 · A.REI.B.4B

Solving quadratics with square roots

You know that a number squared is 49. What could the number have been?

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Quadratic expressions and parabolasGrade 9 What makes an expression quadraticAlgebra 1 Square and cube rootsGrade 8

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Solving quadratics with square roots

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Why this exists

Most steps in solving an equation undo something. Adding 3 is undone by taking 3 away; multiplying by 5 is undone by dividing by 5. Each of those undoing steps gives back exactly one number, because nothing was lost on the way in.

Squaring is different. 7^2 = 49 and (-7)^2 = 49: two different numbers go in and the same number comes out. Squaring forgets the sign. So when an equation says x^2 = 49 and asks you to go backwards, there are two numbers that could have gone in, and an honest answer has to name both. Saying “x is 7” is only half the answer.

The square root symbol alone does not do that. \displaystyle \sqrt{49} means 7 and only 7: it is the principal square root, the one that is not negative. The square root property is the rule that recovers the other one. It is the first tool that solves a quadratic equation without factoring, and it works just as well when the answer is \displaystyle \sqrt{7} as when it is a whole number, which factoring does not.

The method covers every equation in which the unknown appears only once, inside a square. That square may be of a single letter, as in x^2 = 25, or of a whole bracket, as in (x - 3)^2 = 16. Either way the plan is the same: get the square alone on one side, take the root of both sides, and keep both signs. This lesson does the single-letter case first, then the squared bracket in Solving when a binomial is squared, at the end of the Intuition section below.

Definition

The square root property. If a number squared equals k, and k is not negative, then the number is either \displaystyle \sqrt{k} or \displaystyle -\sqrt{k}.

In symbols, for any real number k \ge 0:

x^2 = k \qquad\Longrightarrow\qquad x = \sqrt{k} \quad\text{or}\quad x = -\sqrt{k}

The two answers are written together as \displaystyle x = \pm\sqrt{k}, read “x equals plus or minus the square root of k”. The \pm is shorthand for two separate statements; it never means one number.

The x here can be anything that is squared: a single letter, or a whole expression such as (x - 4). What matters is that the squared thing sits alone on one side, with a number on the other.

Intuition

Draw the graph of y = x^2 and the flat line y = 9. Solving x^2 = 9 means finding every x where the curve reaches height 9.

Figure 1The level y = 9 meets y = x^2 twice, once on each side of the axis.

The curve is a mirror image of itself across the y-axis, because x and -x always have the same square. Whatever height the curve reaches on the right, it reaches at the same height on the left. So a level line above the bottom of the curve crosses it twice, at x = 3 and at x = -3. The two answers are not a trick of the algebra; they are the two sides of the picture.

Now slide the level line up and down.

Figure 2Each flat line is y = k. The sign of k decides how many real solutions x^2 = k has.

Above the bottom of the curve, two crossings. At the bottom itself, where k = 0, the line only touches, at x = 0. Below it, the line misses the curve entirely: no real number has a negative square, so x^2 = -4 has no real solution at all.

Solving when a binomial is squared

The property already solves x^2 = 16: the variable appears once, squared, and the square root undoes the square, giving x = \pm 4.

Now look at

(x - 3)^2 = 16

It looks harder, but it is the same equation wearing a coat. The thing being squared is no longer x but x - 3. Everything that was true of x in x^2 = 16 is true of x - 3 here, so x - 3 = \pm 4. What is left is two one-step linear equations.

When you reach for it. Reach for this move whenever the variable sits inside one squared bracket and appears nowhere else. That happens more often than it seems. It is the shape a quadratic takes after completing the square, it is the shape of vertex form a(x - h)^2 + k, and it is hiding inside any perfect square trinomial such as x^2 - 10x + 25, which is (x - 5)^2 written out.

What you should not do first is multiply the bracket out. (x - 3)^2 = 16 expands to x^2 - 6x - 7 = 0, and now x appears twice and you need a different method to finish. The squared form is the good form. Keep it.

The procedure. The whole idea fits in one line: root both sides, then undo what is left inside.

Take (x - 3)^2 = 16. Call the bracket u for a moment, so u = x - 3 and the equation says u^2 = 16. There are exactly two numbers whose square is 16, namely 4 and -4, so

u = 4 \qquad\text{or}\qquad u = -4

Put x - 3 back in place of u:

x - 3 = 4 \qquad\text{or}\qquad x - 3 = -4

Each of these is a linear equation. Add 3 to both sides of each:

x = 7 \qquad\text{or}\qquad x = -1

Check. (7 - 3)^2 = 4^2 = 16 and (-1 - 3)^2 = (-4)^2 = 16. Both work.

The usual way to write the two cases at once is with \pm, read “plus or minus”:

\begin{aligned} (x - 3)^2 &= 16 \\ x - 3 &= \pm 4 \\ x &= 3 \pm 4 \end{aligned}

and 3 \pm 4 means the two numbers 3 + 4 = 7 and 3 - 4 = -1.

There is also a picture. The equation x - 3 = \pm 4 says that the gap between x and 3 is 4, in one direction or the other. On a number line, that is the two points 4 away from 3:

Figure 3(x - 3)^2 = 16 says x - 3 is 4 or -4, so x sits 4 away from 3 on either side.

So the two solutions always sit evenly either side of the number inside the bracket, here 3. That is the distance reading of the same fact, since \displaystyle \sqrt{(x - 3)^2} = |x - 3|, and |x - 3| = 4 says “x is 4 from 3.”

Why each step is allowed. The first step replaces (\ldots)^2 = 16 by (\ldots) = \pm 4. That is allowed because a number squares to 16 exactly when it is 4 or -4, which is the square root property applied to the bracket, so the two statements are true for exactly the same x. Nothing is gained or lost. The second step adds 3 to both sides of a true equation, which keeps it true and can be undone by subtracting 3, so again the solutions do not change.

Before you take the root, the square must stand alone. The square root property is about an equation of the form “something squared equals a number.” If the equation is 2(x - 3)^2 + 5 = 37, the squared bracket has been multiplied by 2 and then had 5 added to it. Undo those first, in reverse order: subtract 5, then divide by 2. Only then is it “something squared equals a number.”

After the root, undo what is left inside. Once you have x - 3 = \pm 4, what stands between you and x is whatever was done to x inside the bracket. Here it was “subtract 3”, so you add 3. If the bracket were 2x - 3, you would add 3 and then divide by 2.

Think of the whole equation as a set of wrappings around x, put on in order: inside the bracket first, then the square, then whatever is outside it. You take them off in the reverse order, outside first, and the square root sits in the middle of that sequence.

Results

Square root property. For k \ge 0,

x^2 = k \quad\text{exactly when}\quad x = \pm\sqrt{k}

Both values are solutions, and there are no others.

Counting the solutions of x^2 = k by the sign of k:

\begin{aligned} k > 0 &: \quad \text{two real solutions, } \sqrt{k} \text{ and } {-\sqrt{k}} \\ k = 0 &: \quad \text{one real solution, } 0 \\ k < 0 &: \quad \text{no real solution} \end{aligned}

Any squared expression. If u stands for any expression and u^2 = k with k \ge 0, then \displaystyle u = \pm\sqrt{k}. This is what lets the property solve (x - 4)^2 = 49 in one step.

A squared bracket. For k \ge 0,

(x - h)^2 = k \quad\text{exactly when}\quad x = h \pm \sqrt{k}

The two solutions sit the same distance \displaystyle \sqrt{k} either side of h, the number that makes the bracket zero. When k = 0 they meet, and x = h is the only solution. When k < 0 there is no real solution, because the left side is the square of a real number and so is never negative; over the complex numbers the solutions are \displaystyle x = h \pm i\sqrt{-k}.

For instance, (x - 3)^2 = 16 gives x - 3 = \pm 4, so x = 3 \pm 4, that is x = 7 or x = -1.

Why it is true

Suppose k \ge 0. Then \displaystyle \sqrt{k} is a real number, and by the meaning of a square root, \displaystyle (\sqrt{k})^2 = k. Rewrite the equation using that:

\begin{aligned} x^2 &= k \\ x^2 - (\sqrt{k})^2 &= 0 \\ (x - \sqrt{k})(x + \sqrt{k}) &= 0 \end{aligned}

The last line is a difference of squares, factored. A product is zero exactly when one of its factors is zero, which is the zero product property, so

x - \sqrt{k} = 0 \quad\text{or}\quad x + \sqrt{k} = 0

that is, \displaystyle x = \sqrt{k} or \displaystyle x = -\sqrt{k}.

Every line above says the same thing as the line before it, so the argument runs both ways. Both numbers really do solve x^2 = k, and no third number can, because a third solution would have to make one of the two factors zero.

When k < 0 the story is shorter. The square of a real number is never negative: a positive times a positive is positive, a negative times a negative is positive, and 0 \cdot 0 = 0. So no real x has x^2 = k. Over the complex numbers there are two solutions again, \displaystyle x = \pm i\sqrt{-k}, because the imaginary unit was invented precisely so that i^2 = -1.

The squared bracket. Nothing in the argument used the fact that x was a single letter. Put u = x - h; then (x - h)^2 = k is u^2 = k, which holds exactly when \displaystyle u = \sqrt{k} or \displaystyle u = -\sqrt{k}. Putting x - h back for u gives \displaystyle x - h = \pm\sqrt{k}, and adding h to both sides, which can be undone by subtracting h, gives \displaystyle x = h \pm \sqrt{k}. Each step keeps exactly the same solutions, so these two numbers are all of them. When k = 0 both are h, so there is exactly one solution.

When k < 0 the same substitution settles it. For every real x, the number u = x - h is real, and the square of a real number is never negative, so u^2 = k fails; hence no real x solves (x - h)^2 = k. Over the complex numbers \displaystyle u = \pm i\sqrt{-k}, so \displaystyle x = h \pm i\sqrt{-k}.

Worked examples

Example 1

Two perfect squares and one that is not

Solve (a) x^2 = 121 and (b) t^2 = 45.

Solution

(a) The squared term is alone and 121 is positive, so the property applies:

x = \pm\sqrt{121} = \pm 11

So x = 11 or x = -11.

(b) Again the squared term is alone, so \displaystyle t = \pm\sqrt{45}. Now simplify the root: 45 = 9 \cdot 5 and 9 is a perfect square, so

t = \pm\sqrt{9 \cdot 5} = \pm 3\sqrt{5}

Check. \displaystyle (3\sqrt{5})^2 = 9 \cdot 5 = 45, and the negative one squares to the same thing.

Example 2

Isolating the square first

Solve 3y^2 - 7 = 68.

Solution

The property needs the squared term alone with a coefficient of 1, so undo everything around y^2 first, in the reverse order it was done. Add 7 to both sides, which keeps the equation true:

3y^2 = 75

Divide both sides by 3:

y^2 = 25

Only now take roots, keeping both signs:

y = \pm 5

Check. 3(25) - 7 = 75 - 7 = 68. Since 5^2 and (-5)^2 are both 25, one check covers both.

Example 3

A fraction under the root

Solve 2c^2 + 1 = 8.

Solution

Isolate the square: subtract 1, then divide by 2.

\begin{aligned} 2c^2 &= 7 \\ c^2 &= \frac{7}{2} \\ c &= \pm\sqrt{\frac{7}{2}} \end{aligned}

That is a correct answer. It is usually written without a root in the denominator, by rationalizing: split the root, then multiply top and bottom by \displaystyle \sqrt{2}, which is multiplying by 1.

c = \pm\frac{\sqrt{7}}{\sqrt{2}} = \pm\frac{\sqrt{7}\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \pm\frac{\sqrt{14}}{2}

Check. \displaystyle \left(\frac{\sqrt{14}}{2}\right)^2 = \frac{14}{4} = \frac{7}{2}, and \displaystyle 2 \cdot \frac{7}{2} + 1 = 8.

Example 4

When the number is zero or negative

Solve (a) x^2 + 36 = 0 and (b) 7w^2 + 2 = 2.

Solution

(a) Isolate the square: x^2 = -36. The right side is negative, and no real number squares to a negative, so there is no real solution. The graph of y = x^2 never comes down to height -36.

Over the complex numbers, (6i)^2 = 36i^2 = -36, and (-6i)^2 is the same, so

x = \pm 6i

(b) Subtract 2: 7w^2 = 0. Divide by 7: w^2 = 0. The only number whose square is 0 is 0 itself, so w = 0. Here \displaystyle +\sqrt{0} and \displaystyle -\sqrt{0} are the same number, which is why the count drops to one.

Example 5

A squared bracket with whole-number solutions

Solve (x + 5)^2 = 36.

Solution

The square already stands alone, so take the square root of both sides, keeping both signs. Two numbers square to 36, namely 6 and -6:

\begin{aligned} (x + 5)^2 &= 36 \\ x + 5 &= \pm 6 \\ x &= -5 \pm 6 \end{aligned}

So x = -5 + 6 = 1 or x = -5 - 6 = -11. Inside the bracket 5 was added, so the last step subtracts 5.

Check. (1 + 5)^2 = 36 and (-11 + 5)^2 = (-6)^2 = 36.

The two answers sit 6 either side of -5, which is the number that makes the bracket zero. Multiplying out (x + 5)^2 first would have been legal but slower: it puts x in two places, and then the square root property no longer applies. Leaving the square intact is the whole point.

Example 6

A squared bracket with a radical answer

Solve (y - 7)^2 = 20.

Solution

20 is not a perfect square, so the root stays as a radical. Simplify it by taking out the largest square factor, 20 = 4 \cdot 5, so \displaystyle \sqrt{20} = 2\sqrt{5}:

\begin{aligned} (y - 7)^2 &= 20 \\ y - 7 &= \pm\sqrt{20} = \pm 2\sqrt{5} \\ y &= 7 \pm 2\sqrt{5} \end{aligned}

The two solutions are \displaystyle y = 7 + 2\sqrt{5} and \displaystyle y = 7 - 2\sqrt{5}, about 11.47 and 2.53. They are irrational, so this equation does not factor over the integers. The square root method does not care.

Example 7

Isolating the squared bracket first

Solve 3(x - 2)^2 + 5 = 53.

Solution

The squared bracket is not alone: it has been multiplied by 3 and then had 5 added. Take those off in reverse order, subtracting 5 first and then dividing by 3:

\begin{aligned} 3(x - 2)^2 + 5 &= 53 \\ 3(x - 2)^2 &= 48 \\ (x - 2)^2 &= 16 \end{aligned}

Now it is “something squared equals a number,” and the root can be taken:

\begin{aligned} x - 2 &= \pm 4 \\ x &= 2 \pm 4 \end{aligned}

So x = 6 or x = -2.

Check. 3(6 - 2)^2 + 5 = 3 \cdot 16 + 5 = 53, and 3(-2 - 2)^2 + 5 = 3 \cdot 16 + 5 = 53.

Taking the square root earlier would have been wrong: \displaystyle \sqrt{3(x - 2)^2 + 5} is not \displaystyle \sqrt{3}\,(x - 2) + \sqrt{5}. The root undoes a square, and only a square.

Example 8

Fractions on both sides

Solve \displaystyle \left(t - \frac{3}{4}\right)^2 = \frac{5}{16}.

Solution

The square root of a fraction is the root of the top over the root of the bottom, because \displaystyle \left(\frac{\sqrt{5}}{4}\right)^2 = \frac{5}{16}. So \displaystyle \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}:

\begin{aligned} \left(t - \frac{3}{4}\right)^2 &= \frac{5}{16} \\ t - \frac{3}{4} &= \pm\frac{\sqrt{5}}{4} \\ t &= \frac{3}{4} \pm \frac{\sqrt{5}}{4} = \frac{3 \pm \sqrt{5}}{4} \end{aligned}

Because both pieces are over 4, they combine into a single fraction.

Example 9

A square in disguise, with a coefficient inside

Solve 4n^2 + 12n + 9 = 49.

Solution

This does not look like a squared bracket, but the left side is a perfect square trinomial: its first term is (2n)^2, its last is 3^2, and its middle is 2 \cdot 2n \cdot 3 = 12n. So

4n^2 + 12n + 9 = (2n + 3)^2

and the equation is (2n + 3)^2 = 49. Take the root:

2n + 3 = \pm 7

Now undo what is left inside. The bracket did two things to n: multiplied by 2, then added 3. Undo them in reverse, subtracting 3 and then dividing by 2:

\begin{aligned} 2n &= -3 \pm 7 \\ n &= \frac{-3 \pm 7}{2} \end{aligned}

So \displaystyle n = \frac{4}{2} = 2 or \displaystyle n = \frac{-10}{2} = -5.

Check. 4(2)^2 + 12(2) + 9 = 16 + 24 + 9 = 49, and 4(-5)^2 + 12(-5) + 9 = 100 - 60 + 9 = 49.

Here the two solutions sit evenly either side of \displaystyle -\frac{3}{2}, the value that makes 2n + 3 zero, not either side of -3.

Example 10

A squared bracket equal to a negative number

Solve (3v + 1)^2 = -12.

Solution

The square of a real number is never negative. Whatever real v you try, (3v + 1)^2 is 0 or more, so it can never be -12. The equation has no real solutions, and you can see that before doing any arithmetic.

Over the complex numbers it does have two, because \displaystyle \sqrt{-12} = \sqrt{12}\,i = 2\sqrt{3}\,i:

\begin{aligned} 3v + 1 &= \pm 2\sqrt{3}\,i \\ 3v &= -1 \pm 2\sqrt{3}\,i \\ v &= \frac{-1 \pm 2\sqrt{3}\,i}{3} \end{aligned}

Which answer is wanted depends on the question. If it asks for real solutions, the answer is “none.”

Example 11

A length from an area

A square floor tile has an area of 72\text{ cm}^2. How long is each side?

Solution

Call the side s. The area of a square is its side squared, so s^2 = 72. The property gives

s = \pm\sqrt{72} = \pm\sqrt{36 \cdot 2} = \pm 6\sqrt{2}

Both numbers solve the equation, but a length cannot be negative, so the context throws \displaystyle -6\sqrt{2} away. Each side is \displaystyle 6\sqrt{2} \approx 8.49\text{ cm}.

The algebra always hands you both roots. Deciding which ones the situation allows is a separate step, and it belongs to the problem, not to the property.

Example 12

Where a parabola crosses the axis

Where does the graph of y = 2(x - 1)^2 - 18 cross the x-axis?

Solution

The graph crosses the x-axis where y = 0, so solve 2(x - 1)^2 - 18 = 0. Isolate the square, then root:

\begin{aligned} 2(x - 1)^2 &= 18 \\ (x - 1)^2 &= 9 \\ x - 1 &= \pm 3 \\ x &= 1 \pm 3 \end{aligned}

So the graph crosses at x = 4 and x = -2, the points (4, 0) and (-2, 0).

This is why the squared form is worth keeping. The same equation told you, without any work, that the lowest point of the parabola is at x = 1, since that is where the square is 0. The two crossings sit 3 either side of it, which is what symmetry about the line x = 1 requires.

Example 13

Using the property twice

Find every solution of x^4 = 81, real and complex.

Solution

x^4 is (x^2)^2, so this says that the number x^2, squared, gives 81. Apply the property with x^2 as the squared thing:

x^2 = \pm 9

Now there are two equations, and each is a square root problem of its own.

x^2 = 9 gives x = \pm 3.

x^2 = -9 has no real solution, and over the complex numbers gives x = \pm 3i.

So the real solutions are x = 3 and x = -3, and over the complex numbers there are four: 3, -3, 3i and -3i.

Check. 3^4 = 81, and (3i)^4 = 3^4 \cdot i^4 = 81 \cdot 1 = 81.

Boundaries

The squared thing must be alone. \displaystyle \sqrt{a + b} is not \displaystyle \sqrt{a} + \sqrt{b}, so in x^2 + 9 = 25 you cannot take the root of each term: \displaystyle \sqrt{x^2 + 9} is not x + 3. Subtract 9 first, to reach x^2 = 16. In the same way, 2x^2 = 18 must be divided by 2 before the property is used, and in 3(x - 2)^2 + 5 = 53 the 5 and then the 3 come off before the root.

The unknown must appear once, and only inside the square. x^2 + 4x = 5 has x in two places, squared and not, and no single square root undoes both. Rewriting it so that x appears once, inside a square, is exactly what completing the square does. Likewise (x - 3)^2 = x + 1 has x on both sides. Rooting both sides gives \displaystyle x - 3 = \pm\sqrt{x + 1}, which is correct but has not freed x. That equation needs expanding and a different method, such as factoring or the quadratic formula.

The number must not be negative, for real answers. x^2 = k with k < 0 has no real solution. It has two complex ones, \displaystyle \pm i\sqrt{-k}. The same count holds for a squared bracket: once the equation is (\text{bracket})^2 = k, there are two real solutions when k > 0, one when k = 0, none when k < 0 (two complex ones instead, as in Example 10). The bracket changes where the solutions sit, not how many there are. For (x - 3)^2 = k they are \displaystyle 3 \pm \sqrt{k}, which close in on 3 as k shrinks and meet there when k = 0.

Write both signs. \displaystyle \sqrt{16} is 4 only, the principal root. The -4 comes from the equation, not from the symbol, so the \pm is yours to add.

It is about squares, and even powers. A cube keeps its sign: 2^3 = 8 but (-2)^3 = -8, so x^3 = 8 has one real solution, x = 2, and no \pm (see Cube roots and higher roots). A fourth power forgets the sign just as a square does, which is why Example 13 got \pm back.

A root thrown away by the context is still a root. In Example 11, \displaystyle -6\sqrt{2} solves s^2 = 72 perfectly well. The tile rejected it, not the algebra. In a problem about lengths or times, one of the two solutions may be negative; it is still a solution of the equation, but not an answer to the question, and it is dropped for that reason, as in Exercise 41.

Where this goes next

The property is the last step of every quadratic solved by completing the square. That method exists to turn any quadratic, such as x^2 + 6x = 16, into the squared-bracket shape (x + 3)^2 = 25, and from there it is exactly what this lesson does: completing the square is this method with a rewrite in front. Done once with letters instead of numbers, that becomes the derivation of the quadratic formula, which is why the formula carries a \pm in front of its square root: they are the ones from this page.

The shape a(x - h)^2 + k is vertex form. Reading it, you see the lowest or highest point of the parabola at (h, k); solving it equal to 0, as in Example 12, you get the x-intercepts, one on each side of h.

In geometry the property turns up whenever a length is found from a square. The Pythagorean theorem gives c^2 = a^2 + b^2, and the side itself is the positive root. A circle of area A has radius \displaystyle \sqrt{A / \pi}, since \pi r^2 = A.

And the case k < 0, where the real numbers run out, asks for a number whose square is negative. That question is where the complex numbers begin, and it is the one the imaginary unit answers.

Exercises

Solve

x^2 = 36
y^2 = 400
x^2 = 13
a^2 = 48
\displaystyle n^2 = \frac{9}{25}
m^2 = 0

Isolate the square first

x^2 - 64 = 0
5y^2 = 245
2x^2 + 3 = 75
\displaystyle \frac{1}{4}p^2 - 2 = 3
3z^2 = 2
4r^2 - 5 = 0
10 - w^2 = 1

Zero or negative

Say whether each has real solutions. Give the real solutions if there are any, and the complex ones if there are not.

x^2 + 49 = 0
3t^2 + 8 = 8
2x^2 + 50 = 0

A squared bracket

(x + 2)^2 = 81
(x - 9)^2 = 49
(y + 6)^2 = 1
(a + 2)^2 = 100
(m - 1)^2 = 0

Squared brackets with radicals and fractions

(y - 5)^2 = 12
(x - 3)^2 = 28
(b + 5)^2 = 50
\displaystyle \left(r - \frac{1}{2}\right)^2 = \frac{9}{4}
\displaystyle \left(u + \frac{2}{3}\right)^2 = \frac{7}{9}

Isolate the squared bracket first

(x + 1)^2 - 7 = 29
2(p - 4)^2 = 72
5(z + 3)^2 + 2 = 42
-(x - 6)^2 + 20 = 4

A coefficient inside, or a square in disguise

(2x - 3)^2 = 25
(3w + 2)^2 = 18
x^2 + 8x + 16 = 20
9k^2 - 30k + 25 = 4

A squared bracket equal to a negative number

Say whether there are real solutions, and give the complex ones.

(x + 7)^2 = -9
(2s - 1)^2 + 10 = 2

Higher powers

Find every solution of x^4 = 16, real and complex.

In context

A square rug covers 50 square feet. How long is each side, exactly and to two decimal places?
A circle has area 36\pi square meters. Find its radius.
Where does the graph of y = (x - 2)^2 - 9 cross the x-axis?
The side of a square is lengthened by 3 cm, and the new square has an area of 64 cm^2. How long was the side of the original square?

Reasoning

Explain why x^2 = 9 has two solutions but \displaystyle x = \sqrt{9} has only one.
Without solving anything, explain why x^2 = -4 has no real solution.
Prove that if k > 0, then \displaystyle \sqrt{k} and \displaystyle -\sqrt{k} both solve x^2 = k, and that no other real number does.
Explain why the square root property cannot be used straight away on x^2 + 4x = 5, and what has to happen first.
Without solving, explain why the two solutions of (x - h)^2 = k, when k > 0, always have h exactly halfway between them.
Explain why (x - 3)^2 = 16 and (3 - x)^2 = 16 have exactly the same solutions.
Solve (x - 5)^2 = 36 twice: once by square roots, and once by expanding and factoring. Explain why the answers agree, and which route is shorter.
Show that if a \neq 0 and k > 0, the solutions of (ax + b)^2 = k are \displaystyle x = \frac{-b \pm \sqrt{k}}{a}.

Answers

x = \pm 6. 6^2 and (-6)^2 are both 36.
y = \pm 20. \displaystyle \sqrt{400} = 20.
\displaystyle x = \pm\sqrt{13}. 13 has no perfect square factor, so the root stays as it is.
\displaystyle a = \pm 4\sqrt{3}. \displaystyle \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}.
\displaystyle n = \pm\frac{3}{5}. \displaystyle \sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}.
m = 0, the only solution. Only 0 squares to 0.
x = \pm 8. Add 64: x^2 = 64.
y = \pm 7. Divide by 5: y^2 = 49.
x = \pm 6. Subtract 3: 2x^2 = 72. Divide by 2: x^2 = 36.
\displaystyle p = \pm 2\sqrt{5}. Add 2: \displaystyle \frac{1}{4}p^2 = 5. Multiply by 4: p^2 = 20, and \displaystyle \sqrt{20} = 2\sqrt{5}.
\displaystyle z = \pm\frac{\sqrt{6}}{3}. \displaystyle z^2 = \frac{2}{3}, so \displaystyle z = \pm\frac{\sqrt{2}}{\sqrt{3}}; multiply top and bottom by \displaystyle \sqrt{3}.
\displaystyle r = \pm\frac{\sqrt{5}}{2}. 4r^2 = 5, so \displaystyle r^2 = \frac{5}{4} and \displaystyle \sqrt{4} = 2.
w = \pm 3. Subtract 10: -w^2 = -9. Multiply by -1: w^2 = 9.
No real solutions; over the complex numbers, x = \pm 7i. x^2 = -49, and (7i)^2 = -49.
t = 0, one real solution. 3t^2 = 0, so t^2 = 0.
No real solutions; over the complex numbers, x = \pm 5i. 2x^2 = -50, so x^2 = -25.
x = 7 or x = -11. x + 2 = \pm 9, so x = -2 + 9 or x = -2 - 9.
x = 16 or x = 2. x - 9 = \pm 7, so x = 9 \pm 7.
y = -5 or y = -7. y + 6 = \pm 1, so y = -6 \pm 1.
a = 8 or a = -12. a + 2 = \pm 10, so a = -2 \pm 10.
m = 1, the only solution. m - 1 = \pm 0 = 0.
\displaystyle y = 5 \pm 2\sqrt{3}. \displaystyle y - 5 = \pm\sqrt{12} = \pm 2\sqrt{3}.
\displaystyle x = 3 \pm 2\sqrt{7}. \displaystyle x - 3 = \pm\sqrt{28}, and \displaystyle \sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}.
\displaystyle b = -5 \pm 5\sqrt{2}. \displaystyle b + 5 = \pm\sqrt{50}, and \displaystyle \sqrt{50} = 5\sqrt{2}.
r = 2 or r = -1. \displaystyle r - \frac{1}{2} = \pm\frac{3}{2}, so \displaystyle r = \frac{1}{2} \pm \frac{3}{2}.
\displaystyle u = \frac{-2 \pm \sqrt{7}}{3}. \displaystyle u + \frac{2}{3} = \pm\frac{\sqrt{7}}{3}, since \displaystyle \sqrt{9} = 3.
x = 5 or x = -7. Add 7: (x + 1)^2 = 36, so x + 1 = \pm 6.
p = 10 or p = -2. Divide by 2: (p - 4)^2 = 36, so p - 4 = \pm 6.
\displaystyle z = -3 \pm 2\sqrt{2}. Subtract 2: 5(z + 3)^2 = 40. Divide by 5: (z + 3)^2 = 8, and \displaystyle \sqrt{8} = 2\sqrt{2}.
x = 10 or x = 2. Subtract 20: -(x - 6)^2 = -16. Multiply by -1: (x - 6)^2 = 16, so x - 6 = \pm 4.
x = 4 or x = -1. 2x - 3 = \pm 5, so 2x = 3 \pm 5, which is 8 or -2.
\displaystyle w = \frac{-2 \pm 3\sqrt{2}}{3}. \displaystyle 3w + 2 = \pm\sqrt{18} = \pm 3\sqrt{2}, so \displaystyle 3w = -2 \pm 3\sqrt{2}.
\displaystyle x = -4 \pm 2\sqrt{5}. The left side is (x + 4)^2, so (x + 4)^2 = 20 and \displaystyle \sqrt{20} = 2\sqrt{5}.
\displaystyle k = \frac{7}{3} or k = 1. The left side is (3k - 5)^2, so 3k - 5 = \pm 2 and 3k = 5 \pm 2, which is 7 or 3.
No real solutions; over the complex numbers x = -7 \pm 3i. A square cannot be -9, and \displaystyle \sqrt{-9} = 3i.
No real solutions; over the complex numbers \displaystyle s = \frac{1}{2} \pm \sqrt{2}\,i. Subtract 10: (2s - 1)^2 = -8, so \displaystyle 2s - 1 = \pm 2\sqrt{2}\,i and \displaystyle s = \frac{1 \pm 2\sqrt{2}\,i}{2}.
Real: x = \pm 2; complex: x = \pm 2i as well. (x^2)^2 = 16 gives x^2 = \pm 4. x^2 = 4 gives \pm 2, and x^2 = -4 gives \pm 2i.
\displaystyle 5\sqrt{2} feet, about 7.07 feet. s^2 = 50 gives \displaystyle s = \pm 5\sqrt{2}, and a length is positive.
6 meters. \pi r^2 = 36\pi, so r^2 = 36 and r = \pm 6; a radius is positive.
At x = 5 and x = -1. Set y = 0: (x - 2)^2 = 9, so x - 2 = \pm 3.
The original side was 5 cm. If it was s, then (s + 3)^2 = 64, so s + 3 = \pm 8 and s = 5 or s = -11. A length cannot be negative, so -11 is dropped.
x^2 = 9 asks for every number whose square is 9, and both 3 and -3 qualify. The symbol \displaystyle \sqrt{9} names one number, the non-negative square root, which is 3. So \displaystyle x = \sqrt{9} says x = 3 and nothing else.
The square of any real number is positive or zero: a positive times a positive is positive, a negative times a negative is positive, and 0 \cdot 0 = 0. So no real number squares to -4.
First, both work: \displaystyle (\sqrt{k})^2 = k by the meaning of \displaystyle \sqrt{k}, and \displaystyle (-\sqrt{k})^2 = (\sqrt{k})^2 = k. Second, suppose x^2 = k. Then \displaystyle x^2 - (\sqrt{k})^2 = 0, which factors as \displaystyle (x - \sqrt{k})(x + \sqrt{k}) = 0. By the zero product property one factor is zero, so \displaystyle x = \sqrt{k} or \displaystyle x = -\sqrt{k}. No other value is possible.
The unknown appears twice, once squared and once not, so there is no single squared thing standing alone to take the root of. Completing the square first rewrites it as (x + 2)^2 = 9, where x appears once. Then the property gives x + 2 = \pm 3, so x = 1 or x = -5.
The solutions are \displaystyle h + \sqrt{k} and \displaystyle h - \sqrt{k}. Each is the distance \displaystyle \sqrt{k} from h, one on each side, so h is halfway between them. Equivalently, their sum is 2h, so their average is h.
3 - x is the negative of x - 3, and a number and its negative have the same square, so (3 - x)^2 = (x - 3)^2 for every x. The two equations say the same thing, and both have the solutions x = 7 and x = -1. Rooting the second directly gives 3 - x = \pm 4, the same two cases in the other order.
Both routes give x = 11 and x = -1. By roots: x - 5 = \pm 6. By expanding: x^2 - 10x + 25 = 36, so x^2 - 10x - 11 = 0 and (x - 11)(x + 1) = 0. They agree because expanding rewrites the same equation, so it has the same solutions. The root route is shorter, since x appears only once.
Since k > 0, \displaystyle ax + b = \pm\sqrt{k}. Subtract b: \displaystyle ax = -b \pm \sqrt{k}. Since a \neq 0, divide by a: \displaystyle x = \frac{-b \pm \sqrt{k}}{a}.