Chapter 12 of Algebra 1: Concepts and Skills, built for a visual learner. Square root functions and their domains, radicals combined as like terms, radical equations with the extraneous-solution check, rational exponents split into root and power, completing the square drawn as an actual square, and Pythagoras, distance and midpoint on the coordinate grid.
Subject: Algebra 1 · 62 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Algebra 1 · Chapter 12
Square root functions, completing the square, and the geometry that Pythagoras unlocks
Objectives
The final chapter ties the algebra you have built back to geometry, where a lot of it came from in the first place.
Figure (svg): A right triangle with squares drawn on all three sides, the two smaller areas adding to the largest
Section
Section 12.1
Concept
The square root function accepts only non-negative inputs, and grows quickly at first and then more and more slowly.
Figure (svg): The square root curve rising steeply then flattening, starting at the origin and defined only for non-negative inputs
domain restriction — The set of inputs a function will accept. For a square root the expression under the radical must be zero or positive, which is what stops the curve to the left.
Worked example
Find the domain of the function below, then describe its graph.
\[ y = \sqrt{x - 3} \]
Set the expression under the radical to be non-negative
Why: A square root of a negative number is not a real number, so the inside must be zero or more.
\[ x - 3 \ge 0 \;\Longrightarrow\; x \ge 3 \]
Describe the graph from that
Why: The curve begins at x equal to 3, where the output is zero, and rises to the right in the usual square root shape.
Figure (svg): The square root curve shifted three units right, beginning at the point three comma zero
Verify: test a value just inside and just outside the domain
Why: At x equal to 4 the inside is 1 and the output is 1, which is fine. At x equal to 2 the inside is negative 1, which has no real root — so the domain boundary is correct.
Prediction
Commit before graphing.
\[ y = \sqrt{x + 5} \]
Predict first
What is the domain of this function?
Correct: x is at least negative 5.
Why: The inside must be zero or positive, so x plus 5 is at least zero, giving x at least negative 5. Adding inside the radical shifts the curve LEFT, which feels backwards — the inside reaches zero sooner, so the starting point moves earlier.
Sorting
For the function shown, sort each input.
\[ y = \sqrt{2x - 6} \]
Sort into buckets
Which inputs does this function accept?
Tweak it
The shape never changes. Only where it begins does.
Parameter explorer
Drag the shift. Which way does the curve move, and where does its starting point land?
\[ y = \sqrt{x - {h}} \]
Section
Section 12.2
Concept
Two radical terms add only when the expression under the sign is identical — exactly the like-terms rule from Chapter 2.
Figure (svg): Radical terms sorted like algebra tiles: only matching radicands can be combined
\[ 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2} \]
Worked example
Simplify the expression below.
\[ \sqrt{12} + \sqrt{27} \]
Simplify each radical first
Why: They look unlike, but simplifying may reveal a shared radicand. Twelve is 4 times 3 and 27 is 9 times 3.
\[ 2\sqrt{3} + 3\sqrt{3} \]
Now combine, because the radicands match
Why: Two of something plus three of the same something is five of it.
\[ 5\sqrt{3} \]
Figure (svg): Two unlike radicals each simplifying to a multiple of root three, then combining
Verify: check numerically
Why: The root of 12 is about 3.46 and the root of 27 is about 5.20, summing to about 8.66. Five times the root of 3 is about 8.66 as well.
Error analysis
Find the illegal step.
Annotate
On: \( \sqrt{9} + \sqrt{16} \;\overset{?}{=}\; \sqrt{25} = 5 \)
A radical over a sum must have the sum finished first; two separate radicals stay separate.
Discrimination
Simplify mentally first, then decide.
Sort into buckets
Which pairs can be combined into a single term?
Fill the middle
Fill each blank.
Fill in the blanks
\sqrt5 - \sqrt2 = 3\sqrt___ - ___\sqrt___ = ___\sqrt___
Why: Fifty splits as 25 times 2, so its root is 5 root 2, and 8 splits as 4 times 2, giving 2 root 2. Now the radicands match and the coefficients subtract to 3. Neither radical could be touched before simplifying, which is why simplify-first is the rule here.
Explain it
A classmate is surprised that 3 root 2 plus 5 root 2 is 8 root 2.
Discussion prompt
Explain the connection to like terms, without using the word radicand.
Hint: What would you say if the radical were replaced by the letter x?
Answer:
Root 2 is just some fixed number, so three of it plus five of it is eight of it — exactly as three apples plus five apples is eight apples.
And that is why root 2 plus root 3 does not simplify: they are different numbers, so you are adding two unlike things and there is no single count to report.
Section
Section 12.3
Concept
To remove a radical, square both sides. But squaring is not reversible, so it can create solutions that were never there.
Figure (svg): A solving chain where squaring both sides produces two candidates, one of which fails the check
Every radical equation therefore ends with a check against the original — not against a later line.
Worked example
Solve the equation below.
\[ \sqrt{x + 6} = x \]
Square both sides
Why: Squaring removes the radical and leaves an ordinary quadratic.
\[ x + 6 = x^2 \]
Rearrange and solve the quadratic
Why: Set it to zero and factor.
\[ x^2 - x - 6 = 0 \;\Longrightarrow\; (x - 3)(x + 2) = 0 \]
\[ x = 3 \quad \text{or} \quad x = -2 \]
Check both against the original
Why: At x equal to 3 the left is the root of 9, which is 3, matching. At x equal to negative 2 the left is the root of 4, which is positive 2, and that does not equal negative 2.
\[ x = 3 \quad \text{only} \]
Figure (svg): A solving chain where squaring both sides produces two candidates, one of which fails the check
Verify: state why the second answer failed
Why: A radical always produces a non-negative result, so it can never equal a negative number. The value negative 2 was created by the squaring step and is extraneous.
Anomaly
The algebra was flawless and one answer is still wrong.
\[ \sqrt{x} = -3 \]
Predict first
Squaring gives x equals 9. Why is that not a solution?
Correct: Because a radical never produces a negative result, so the original equation is impossible.
\[ \sqrt{x} \ge 0 \text{ always, so it cannot equal } -3 \]
Why: Squaring both sides destroys sign information: both 3 and negative 3 square to 9, so squaring turns an impossible equation into a solvable one. The original demanded a non-negative quantity equal a negative number, which nothing can do. The equation genuinely has no solution.
Fill the middle
Fill each blank.
Fill in the blanks
\sqrt25 = 5 \;\Longrightarrow\; 2x + 1 = 12 \;\Longrightarrow\; x = ___
Why: Squaring both sides gives 2x plus 1 equals 25, so 2x is 24 and x is 12. Checking: 2 times 12 plus 1 is 25, whose root is 5, matching the right side. Here the right side was already positive, so no extraneous solution could appear — but the check still costs nothing.
Explain it to yourself
Solving a linear equation needs no check. Solving a radical one does. Say why.
Discussion prompt
What is different about squaring that makes the check compulsory?
Hint: Can you undo squaring and get back exactly where you started?
Answer:
Every step in a linear solve is reversible: whatever you did, you could undo. So a solution of the last line is automatically a solution of the first.
Squaring is not reversible, because two different numbers square to the same thing. So the squared equation can be true where the original was false, and only substituting back into the original can tell the two apart.
Sorting
For the equation shown, sort each candidate the algebra produced.
\[ \sqrt{x + 2} = x \]
Sort into buckets
Which candidates are genuine solutions?
Section
Section 12.4
Concept
An exponent of one half means a square root, one third means a cube root, and so on. The rules from Chapter 8 all still apply.
\[ x^{1/2} = \sqrt{x} \qquad x^{2/3} = \sqrt[3]{x^2} \]
Figure (svg): A fractional exponent split into its two jobs: the denominator is the root and the numerator is the power
Worked example
Evaluate each expression below.
\[ 16^{1/2} \qquad 8^{2/3} \]
For the first, read the exponent as a square root
Why: A denominator of 2 means a square root, and 16 has root 4.
\[ 16^{1/2} = 4 \]
For the second, take the root first, then the power
Why: The denominator 3 says cube root, and 8 has cube root 2. Then the numerator 2 says square it.
\[ 8^{2/3} = (2)^2 = 4 \]
Note why the root goes first
Why: Taking the root first keeps the numbers small. Doing the power first gives 64, whose cube root is also 4 — same answer, harder arithmetic.
Figure (svg): Two routes through eight to the two thirds, both arriving at four, with the root-first route marked as easier
Verify: check by rewriting as a radical
Why: Eight to the two thirds is the cube root of 8 squared, which is the cube root of 64, which is 4 — matching both routes.
Matching
Read the denominator as the root and the numerator as the power.
Match the pairs
Why: In each case take the root indicated by the denominator, then raise to the power in the numerator. Four to the three halves is the square root of 4, which is 2, cubed to give 8. Sixteen to the three quarters is the fourth root of 16, which is 2, cubed to give 8 as well.
Notation
One symbol doing two jobs at once.
Annotate
On: \( x^{3/4} \)
Bottom is root, top is power. Every rational-exponent question is that one sentence applied carefully.
Prediction
Commit before computing.
\[ 9^{1/2} \quad \text{versus} \quad 9^{2} \]
Predict first
Which is larger?
Correct: 9 squared, by a very long way — 81 against 3.
Why: An exponent bigger than one makes a number grow, and an exponent between zero and one makes it shrink. Nine to the one half is the square root, which is 3, while 9 squared is 81. Fractional exponents shrinking a number surprises people, but it follows directly from a root being the opposite of a power.
Section
Section 12.5
Concept
Completing the square turns any quadratic into a perfect square plus a constant, by adding exactly the piece the square is missing.
Figure (svg): An area model showing a square being completed by adding the missing corner piece
\[ x^2 + 6x + 9 = (x + 3)^2 \]
Worked example
Solve the equation below by completing the square.
\[ x^2 + 6x - 7 = 0 \]
Move the constant to the other side
Why: Leave only the x terms on the left, so there is room for the corner piece.
\[ x^2 + 6x = 7 \]
Halve the middle coefficient and square it
Why: Half of 6 is 3, and 3 squared is 9. That is the missing corner.
Add it to both sides
Why: Adding to one side only would break the equation, so the 9 goes on both.
\[ x^2 + 6x + 9 = 16 \]
Write the left as a perfect square and take roots
Why: The left is now the square of x plus 3, and the right is 16.
\[ (x + 3)^2 = 16 \;\Longrightarrow\; x + 3 = \pm 4 \]
\[ x = 1 \quad \text{or} \quad x = -7 \]
Figure (svg): An area model showing a square being completed by adding the missing corner piece
Verify: substitute both roots into the original
Why: At x equal to 1: 1 plus 6 minus 7 is 0. At x equal to negative 7: 49 minus 42 minus 7 is 0. Both work.
Fill the middle
Halve the middle coefficient, then square it.
Fill in the blanks
x^2 + 10x + 25 = (x + 5)^2
Why: Half of 10 is 5, and 5 squared is 25. So adding 25 completes the square, and the result factors as x plus 5 all squared. The number inside the bracket is always half the middle coefficient, which is a useful check on the whole process.
Prediction
Commit before computing.
\[ x^2 - 8x + \underline{\;\;\;} \]
Predict first
What constant completes the square?
Correct: 16.
Why: Half of negative 8 is negative 4, and negative 4 squared is positive 16. The result is the square of x minus 4. The answer is always positive because it is a square, even when the middle coefficient is negative — answering negative 16 means the squaring step was skipped.
Socratic
The quadratic formula already solves everything. So why learn this?
Discussion prompt
What does completing the square give you that the quadratic formula does not?
Hint: What can you read off (x + 3)^2 - 16 that you cannot read off the original?
Answer:
It puts the quadratic into vertex form, which hands you the turning point directly — something the formula never shows you.
It is also where the quadratic formula comes from: completing the square on the general equation a x squared plus b x plus c produces the formula itself. Learning it is learning why the formula is true rather than just that it is.
Section
Section 12.6
Concept
In a right triangle, the squares on the two shorter sides together have exactly the same area as the square on the longest side.
Figure (svg): A right triangle with squares drawn on all three sides, the two smaller areas adding to the largest
\[ a^2 + b^2 = c^2 \]
Worked example
A right triangle has legs of 5 and 12. Find the hypotenuse.
Identify which side is the hypotenuse
Why: It is always the side opposite the right angle, and always the longest. Getting this wrong is the main source of error.
Substitute into the theorem
Why: The two legs are the a and b; the unknown is c.
\[ 5^2 + 12^2 = c^2 \]
\[ 25 + 144 = 169 \]
Take the square root
Why: The side length is a distance, so only the positive root is meaningful here.
\[ c = 13 \]
Figure (svg): A right triangle with legs five and twelve and hypotenuse thirteen
Verify: check the arithmetic against the theorem
Why: Thirteen squared is 169, and 25 plus 144 is also 169, so the three sides genuinely satisfy the relationship.
Error analysis
A right triangle has a hypotenuse of 10 and one leg of 6. Find the other leg.
Annotate
On: \( 10^2 + 6^2 = c^2 \;\overset{?}{\Longrightarrow}\; c = \sqrt{136} \approx 11.7 \)
The hypotenuse is always the largest number in the equation. If your answer beats it, the setup was wrong.
Sorting
The converse of Pythagoras: if the three sides satisfy the relationship, the triangle must have a right angle.
Sort into buckets
Which of these side triples form a right triangle?
Real world
A ladder 13 feet long leans against a wall with its foot 5 feet from the base.
Discussion prompt
How far up the wall does it reach, and which side is the hypotenuse here?
Hint: Which of the three lengths is opposite the right angle?
Answer:
The ladder itself is the hypotenuse, because it is opposite the right angle formed by the wall and the ground.
\[ 5^2 + h^2 = 13^2 \;\Longrightarrow\; h^2 = 144 \;\Longrightarrow\; h = 12 \]
It reaches 12 feet up. The size check works: the height must be less than the ladder's length, and 12 is less than 13.
Section
Section 12.7
Concept
Any two points on a plane form a right triangle with the grid. The distance formula is Pythagoras applied to it.
Figure (svg): Two points on a plane with a right triangle drawn between them, the hypotenuse being the distance
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Worked example
Find the distance between the two points below.
\[ (1, 1) \text{ and } (5, 4) \]
Find the horizontal and vertical gaps
Why: These are the two legs of the triangle: 5 minus 1 is 4 across, and 4 minus 1 is 3 up.
Square and add them
Why: This is exactly the Pythagorean step, with the gaps as the legs.
\[ 4^2 + 3^2 = 16 + 9 = 25 \]
Take the square root
Why: The distance is the hypotenuse.
\[ d = 5 \]
Figure (svg): Two points on a plane with a right triangle drawn between them, the hypotenuse being the distance
Verify: check the order of subtraction does not matter
Why: Subtracting the other way gives negative 4 and negative 3, and squaring removes both signs, so the answer is unchanged. Distance cannot depend on which point you start from.
Explain it to yourself
You can subtract the coordinates either way round and get the same distance. Say why.
Discussion prompt
Why does swapping which point you call first leave the distance unchanged?
Hint: What does squaring do to a negative number?
Answer:
Because each difference gets squared, and squaring destroys the sign. Four and negative four both square to 16.
That matches common sense: the distance from your house to school is the same as from school to your house. A formula that gave two different answers would be describing something other than distance.
Prediction
Commit before computing.
\[ (-2, 3) \text{ and } (1, -1) \]
Predict first
What is the distance between these points?
Correct: 5.
Why: The horizontal gap is 1 minus negative 2, which is 3, and the vertical gap is negative 1 minus 3, which is negative 4. Squaring gives 9 plus 16, which is 25, and the root is 5. Answering 7 means the gaps were added rather than squared and added — that would be the walking distance along the grid, not the straight-line distance.
Section
Section 12.8
Concept
The midpoint of a segment is found by averaging the x coordinates and averaging the y coordinates separately.
\[ M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right) \]
Figure (svg): Two points on a plane with their midpoint marked exactly halfway along the joining segment
Worked example
Find the midpoint of the segment joining the two points, then find the missing endpoint of another segment.
\[ (2, -3) \text{ and } (8, 5) \]
Average the x coordinates
Why: Two plus 8 is 10, halved is 5.
Average the y coordinates
Why: Negative 3 plus 5 is 2, halved is 1.
\[ M = (5, 1) \]
Now work backwards: if one endpoint is (1, 4) and the midpoint is (3, 7), find the other
Why: The midpoint is the average, so the other endpoint must be as far past the midpoint as the first one is before it.
\[ (5, 10) \]
Figure (svg): A segment with one endpoint and the midpoint given, and the equal step continued to find the far endpoint
Verify: average the recovered endpoints
Why: One plus 5 is 6, halved is 3, and 4 plus 10 is 14, halved is 7 — giving the midpoint (3, 7) we were told. The recovered endpoint is correct.
Comparison
Fill the blanks. These two formulas are easy to confuse and do opposite kinds of job.
Comparison matrix
| distance | midpoint | |
|---|---|---|
| what it returns | a single number, a length | a point, with two coordinates |
| the operation | subtract, square, add, root | add and halve |
| does order matter | no, squaring removes the sign | no, addition is commutative |
The quickest way to keep them apart: a distance is one number, a midpoint is a pair. If your answer has the wrong shape, you used the wrong formula.
Section
Section 12.9
Concept
Every worked example in this course has been a small proof: each step justified by a rule, chained until the conclusion is unavoidable.
Figure (svg): A chain of four linked steps, each labelled with the rule that justifies it
Worked example
Prove that the sum of two even numbers is always even.
Write down what an even number means, in symbols
Why: An even number is two times some whole number. Naming the general case is what makes the proof cover all of them at once.
\[ a = 2m \qquad b = 2n \]
Add them
Why: This is ordinary algebra, and it is the only computation in the whole proof.
\[ a + b = 2m + 2n \]
Factor out the shared 2
Why: Factoring reveals the structure the conclusion needs.
\[ a + b = 2(m + n) \]
Read the conclusion off the form
Why: The result is two times a whole number, which is exactly the definition of even. Since m and n stood for any whole numbers, this covers every case.
Figure (svg): Two even bars each made of pairs, joined into one longer bar still made entirely of pairs
Verify: test the claim on a specific pair
Why: Six and four are both even, and their sum is 10, which is even. That confirms the proof rather than replacing it — one example could never establish the general claim.
Counterexample
A classmate claims: the sum of two odd numbers is always odd.
Discussion prompt
Disprove it, and say why one example is enough here when one example never proves a general claim.
Hint: Try the two smallest odd numbers you can think of.
Answer:
Three plus five is 8, which is even. The claim is false.
A single counterexample is enough to disprove a universal claim, because the claim asserted something about every pair. Proving one requires covering all cases; disproving one requires only finding a single failure.
This asymmetry is worth keeping: proof needs algebra, disproof often needs only a well-chosen example.
Ranking
Put these steps of the even-plus-even proof into a valid order.
Put in order
Why: Definitions come first, because nothing can be argued about undefined objects. Then the computation, then the rearrangement that exposes the structure, and only then the conclusion. Stating the conclusion before the factoring would be assuming what you set out to show.
Pattern
Nine sections, and each reduces to one governing move.
Trap
Solve the equation below.
\[ \sqrt{2x + 3} = x \]
Square both sides, solve, and report both answers
Why: The quadratic gives x equal to 3 and x equal to negative 1, and both look like perfectly ordinary solutions.
\[ x = 3 \quad \text{and} \quad x = -1 \]
Substituting negative 1 gives the root of 1, which is positive 1, not negative 1. One of the answers is a fabrication.
Solve the same equation, then check both answers against the original.
Square, solve, then substitute each candidate back
Why: The check is a step of the method, not an optional courtesy. A radical is never negative, so any candidate making the right side negative fails immediately.
\[ x = 3 \quad \text{only} \]
Testing 3: the root of 9 is 3, matching. The other candidate was created by the squaring and is extraneous.
Check
Solve it on paper before you click.
Check your understanding
Simplify the square root of 45 plus the square root of 20.
Answer: A
Why: Root 45 is 3 root 5 and root 20 is 2 root 5. Once both are simplified the radicands match, so they combine to 5 root 5. Numerically that is about 11.18, and the two original roots sum to about 6.71 plus 4.47, which agrees.
Check
Solve it on paper before you click.
Check your understanding
What is the distance between (-1, 2) and (3, 5)?
Answer: A
Why: The horizontal gap is 3 minus negative 1, which is 4, and the vertical gap is 5 minus 2, which is 3. Squaring and adding gives 16 plus 9, which is 25, and the square root is 5.
Commit first
Answer, then rate your confidence.
\[ x^2 - 10x = 11 \]
Predict first
What must you add to both sides to complete the square?
Correct: 25.
\[ (x - 5)^2 = 36 \;\Longrightarrow\; x = 11 \text{ or } x = -1 \]
Why: Half of negative 10 is negative 5, and negative 5 squared is positive 25. Adding 25 to both sides gives the square of x minus 5 equals 36, so x minus 5 is plus or minus 6, giving x equal to 11 or negative 1. The answer is always positive because it is a square.
Exit ticket
The last commitment of the whole course.
Predict first
Which of these is shakiest right now?
Correct: Whatever you picked is the one to drill first.
Why: The second one matters most in Algebra 2, where radical and rational equations appear constantly and unchecked answers cost whole questions. The habit is the same one this whole course has been building: substitute your answer back into the original problem, every time.
Connect it up
Not just this chapter — the whole year, on one page.
Draw it
Down the left, list the twelve chapters. Draw an arrow from each chapter to every later one that uses it. Circle the three ideas that appear in the most arrows. Then write one sentence on what you would tell yourself at the start of Chapter 1, knowing what you know now.
Substituting to check, doing the same thing to both sides, and drawing the picture before the algebra — those three show up in nearly every chapter, and they are what carries into Algebra 2.
Recap
This is the last chapter of the course, and it closes the loop back to the geometry algebra came from.
| if you remember one thing | it should be |
|---|---|
| about radicals | combine only matching radicands, and simplify before deciding |
| about squaring | it is not reversible, so always check against the original |
| about completing the square | half the middle coefficient, squared, added to both sides |
| about the whole course | substitute your answer back in — it catches almost everything |
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