The product and quotient properties of square roots, the two conditions for a radical expression to be simplified, rationalising denominators using conjugates, solving quadratic equations by taking square roots, and the dropped-object height model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Solve Quadratic Equations by Finding Square Roots
Objectives
Five outcomes. The fourth is the point; the first three make its answers presentable.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-269 — the lesson these objectives are drawn from
Warm-up
Lessons 4.3 and 4.4 solved quadratics by factoring. That method needs the quadratic to factor, and most do not.
Discussion prompt
Try to solve 3x squared plus 5 equals 41 by factoring. Write it in standard form first, then look for the two brackets.
Hint: Get it to standard form and see what the constant term is.
Answer:
\[ 3x^2 + 5 = 41 \;\Longrightarrow\; 3x^2 - 36 = 0 \;\Longrightarrow\; 3(x^2 - 12) = 0 \]
Twelve is not a perfect square, so x squared minus 12 does not factor over the integers and the method stops. But the equation is easy in a different way: it says a square equals 12, and you can simply ask what squares to 12. That is this lesson.
Concept
If a quadratic equation has no first-power term, x can be isolated inside a single square and the square undone directly. Because both a positive and a negative number square to the same value, undoing a square always produces two answers.
radical — An expression of the form root s. The symbol is the radical sign and the number s beneath it is the radicand. A positive number has two square roots, and the positive one is called the principal square root.
\[ x^2 = s \;\Longrightarrow\; x = \pm\sqrt{s} \qquad (s > 0) \]
The method is narrower than factoring in one way and wider in another: it needs the equation to be a square equal to a number, but it does not care whether the answer is rational.
Figure (svg): The parabola y equals x squared crossed by the horizontal line y equals twelve at two points
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-267
Section
Section 1
Concept
For non-negative a and b, the square root of a product is the product of the square roots, and the square root of a quotient is the quotient of the square roots. Read left to right, both properties break a radical into pieces, which is how radicals get simplified.
\[ \sqrt{ab} = \sqrt{a}\cdot\sqrt{b}, \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \]
An expression counts as simplified only when no radicand has a perfect-square factor other than 1 and no radical sits in a denominator. Both conditions are part of the definition, so an answer failing either is unfinished.
Figure (svg): The product and quotient properties of square roots, each beside a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-266 — Properties of Square Roots
Picture it
Two clauses, each with an expression that fails it and one that passes.
Figure (svg): The two conditions an expression must meet to count as simplified, with an example failing and then passing each
The first clause is handled by the product property, splitting off the largest perfect-square factor. The second is handled by rationalising, which is the next idea.
Worked example
Example 1, all four parts.
\[ \text{Simplify } \sqrt{80}, \; \sqrt{6}\cdot\sqrt{21}, \; \sqrt{\tfrac{4}{81}}, \; \sqrt{\tfrac{7}{16}}. \]
Split off the largest square factor of 80
Why: Eighty is 16 times 5, and 16 is a perfect square, so the root becomes 4 times the root of 5.
\[ \sqrt{80} = 4 \sqrt{5} \]
Combine the second pair first
Why: The product property run backwards gives the root of 126, and 126 is 9 times 14.
\[ \sqrt{126} = 3 \sqrt{14} \]
Split the third as a quotient
Why: Both 4 and 81 are perfect squares, so the whole expression becomes a fraction of whole numbers.
\[ \sqrt{\frac{4}{81}} = \frac{2}{9} \]
Split the fourth the same way
Why: Sixteen is a perfect square but 7 is not, so only the denominator resolves.
\[ \sqrt{\frac{7}{16}} = \sqrt{7} / 4 \]
Figure (svg): The solution to Worked example use the properties shown as a ladder of expressions, one row per algebraic move
\[ 4\sqrt{5}, \quad 3\sqrt{14}, \quad \tfrac{2}{9}, \quad \tfrac{\sqrt{7}}{4} \]
Verify: square one answer back
Why: Four root 5, squared, is 16 times 5, which is 80 — the original radicand. Note the second part used the product property in both directions: first to combine two radicals into one, then to split a square factor back out. That is the usual pattern for a product of radicals.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-266
Matching
Find the largest perfect-square factor of each radicand.
Match the pairs
Why: Each radicand splits as a perfect square times a number with no square factors left. The last one has to be combined into a single radical first, since 10 and 15 individually have no square factors but their product, 150, contains 25.
Worked example
Guided Practice 1 to 8.
\[ \text{Simplify } \sqrt{27}, \sqrt{98}, \sqrt{10}\cdot\sqrt{15}, \sqrt{8}\cdot\sqrt{28}, \sqrt{\tfrac{9}{64}}, \sqrt{\tfrac{15}{4}}, \sqrt{\tfrac{11}{25}}, \sqrt{\tfrac{36}{49}}. \]
The first two
Why: Twenty-seven is 9 times 3 and 98 is 49 times 2, and in each case the square factor comes out.
\[ 3 \sqrt{3}; 7 \sqrt{2} \]
The two products
Why: Ten times 15 is 150, which is 25 times 6. Eight times 28 is 224, which is 16 times 14.
\[ 5 \sqrt{6}; 4 \sqrt{14} \]
The two quotients that resolve fully
Why: Nine over 64 and 36 over 49 have perfect squares top and bottom.
\[ \frac{3}{8}; \frac{6}{7} \]
The two that only partly resolve
Why: Fifteen over 4 and 11 over 25 have perfect-square denominators only, so a radical survives in the numerator.
\[ \sqrt{15} / 2; \sqrt{11} / 5 \]
Figure (svg): The solution to Worked example eight more shown as a ladder of expressions, one row per algebraic move
\[ 3\sqrt{3}, \; 7\sqrt{2}, \; 5\sqrt{6}, \; 4\sqrt{14}, \; \tfrac{3}{8}, \; \tfrac{\sqrt{15}}{2}, \; \tfrac{\sqrt{11}}{5}, \; \tfrac{6}{7} \]
Verify: check the largest square factor was taken
Why: For 224 the factor 16 was used; had 4 been used instead, the result would be 2 root 56, which is not simplified because 56 still contains the square factor 4. Taking the largest square factor at once avoids a second pass, and the test for being finished is always the same: does the radicand still have a square factor?
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-266
Trap
\[ \sqrt{9 + 16} \]
Split the radical over the two terms
Why: The product property is applied to a sum instead of a product.
\[ \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \quad \text{(wrong)} \]
But 9 plus 16 is 25, and the root of 25 is 5, not 7. The two answers are not even close.
\[ \sqrt{9 + 16} = \sqrt{25} = 5 \]
Evaluate inside the radical first, then take the root
Why: Radicals distribute over multiplication and division only, never over addition or subtraction.
\[ \sqrt{ab} = \sqrt{a}\sqrt{b} \quad \text{but} \quad \sqrt{a + b} \neq \sqrt{a} + \sqrt{b} \]
The same restriction will govern every operation in Chapter 6. A quick numerical test on small numbers, as here, settles any doubt about whether a rule distributes.
Sorting
Test both clauses of the definition.
Sort into buckets
Sort each expression.
The last item is the instructive one: partially simplifying is easy to mistake for finishing, so always re-test the radicand you end up with.
Fill the middle
Guided Practice 4.
Fill in the blanks
\sqrt4\cdot\sqrt___ = \sqrt___ = \sqrt___\cdot\sqrt___ = ___\sqrt___
Why: Combining first gives 224, whose largest square factor is 16, leaving 4 root 14. Fourteen is 2 times 7, neither repeated, so nothing more can come out. Combining before splitting is usually the faster route for a product of two radicals.
Prediction
Commit before reasoning.
Predict first
How many real square roots does 9 have, and what does the radical symbol denote?
Correct: Two roots, 3 and -3, and the symbol denotes only the positive one.
\[ \sqrt{9} = 3 \quad \text{but} \quad x^2 = 9 \;\Longrightarrow\; x = \pm 3 \]
Why: Both 3 squared and negative 3 squared are 9, so 9 has two square roots. The radical symbol is defined to give the principal, or positive, root, which is why the symbol alone returns 3. That is exactly why solving x squared equals 9 requires writing plus-or-minus explicitly: the symbol will not supply the negative root by itself.
Section
Section 2
Concept
A radical in a denominator breaks the second clause of the definition of simplified. Multiplying the numerator and denominator by a suitable expression clears it: by the radical itself when the denominator is a single radical, and by the conjugate when the denominator is a rational number plus or minus a radical.
conjugates — The expressions a plus root b and a minus root b. Their product, a squared minus b, is always rational, which is what makes them useful.
\[ (a + \sqrt{b})(a - \sqrt{b}) = a^2 - b \]
Multiplying top and bottom by the same expression does not change the value of the fraction, only the way it is written — which is why rationalising is always legal.
Figure (svg): A table matching each form of denominator with the expression that clears it
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 267-267 — Rationalizing the Denominator
Picture it
Three forms of denominator, three things to multiply by.
Figure (svg): A table matching each form of denominator with the expression that clears it
The bottom row explains all three: the cross terms cancel, leaving a difference of squares in which the radical has been squared away.
Worked example
Example 2. One single radical, one conjugate.
\[ \text{Simplify } \sqrt{\tfrac{5}{2}} \text{ and } \frac{3}{7 + \sqrt{2}}. \]
Split the first with the quotient property
Why: The root of 5 over 2 becomes the root of 5 over the root of 2, which has a radical below.
\[ \sqrt{5} / \sqrt{2} \]
Multiply top and bottom by the root of 2
Why: The denominator becomes the root of 2 squared, which is 2, and the numerator becomes the root of 10.
\[ \sqrt{10} / 2 \]
Identify the conjugate of the second denominator
Why: The denominator is 7 plus the root of 2, so its conjugate is 7 minus the root of 2.
\[ \text{multiply by } \frac{7 - \sqrt{2}}{7 - \sqrt{2}} \]
Multiply out and simplify
Why: The numerator is 21 minus 3 root 2; the denominator is 49 minus 7 root 2 plus 7 root 2 minus 2, and the middle terms cancel, leaving 47.
\[ \frac{21 - 3 \sqrt{2}}{47} \]
Figure (svg): The solution to Worked example rationalise two denominators shown as a ladder of expressions, one row per algebraic move
\[ \frac{\sqrt{10}}{2} \qquad \text{and} \qquad \frac{21 - 3\sqrt{2}}{47} \]
Verify: check the value is unchanged
Why: The root of 5 over the root of 2 is about 1.581, and the root of 10 over 2 is about 1.581 as well. Multiplying by a fraction equal to 1 cannot change a value, so any rationalisation is guaranteed to preserve it — the check is really on the arithmetic, not on the legality.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 267-267
Matching
Read the form first.
Match the pairs
Why: The two single radicals are cleared by themselves; the two sums and differences are cleared by their conjugates. The last one shows a useful order: simplifying the radical before rationalising leaves smaller numbers to multiply.
Worked example
Guided Practice 13 to 16.
\[ \text{Simplify } \frac{26}{7 - \sqrt{5}}, \; \frac{2}{4 + \sqrt{11}}, \; \frac{21}{9 + \sqrt{7}}, \; \frac{4}{8 - \sqrt{3}}. \]
First: multiply by 7 plus root 5
Why: The denominator becomes 49 minus 5, which is 44, and the numerator becomes 26 times the quantity 7 plus root 5.
\[ 26(7 + \sqrt{5}) / 44 \]
Reduce the first
Why: Twenty-six and 44 share a factor of 2.
\[ 13(7 + \sqrt{5}) / 22 \]
Second and third
Why: Sixteen minus 11 is 5, and 81 minus 7 is 74, giving the two denominators.
\[ 2(4 - \sqrt{11}) / 5; 21(9 - \sqrt{7}) / 74 \]
Fourth
Why: Sixty-four minus 3 is 61, which shares no factor with 4.
\[ 4(8 + \sqrt{3}) / 61 \]
Figure (svg): The solution to Worked example four conjugate problems shown as a ladder of expressions, one row per algebraic move
\[ \frac{13(7+\sqrt{5})}{22}, \; \frac{2(4-\sqrt{11})}{5}, \; \frac{21(9-\sqrt{7})}{74}, \; \frac{4(8+\sqrt{3})}{61} \]
Verify: check one denominator numerically
Why: For the first, 7 minus root 5 is about 4.764, and 26 divided by that is about 5.458. The answer 13 times the quantity 7 plus root 5, over 22, is 13 times 9.236 over 22, which is also about 5.458. Reducing the fraction at the end is easy to forget and does not affect correctness, only the form.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-268
Error analysis
A student tries to rationalise a denominator that is a sum.
Annotate
On: \( \frac{3}{7 + \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{7\sqrt{2} + 2} \)
Read the form of the denominator before choosing what to multiply by. The whole point of the table is that the three forms need three different multipliers.
Fill the middle
Example 2b, at the denominator.
Fill in the blanks
(7 + \sqrt47)(7 - \sqrt___) = 49 - 7\sqrt___ + 7\sqrt___ - 2 = ___
Why: The two cross terms are exact opposites and cancel, leaving 49 minus 2, which is 47. That cancellation is the entire reason conjugates work, and it is the difference-of-squares pattern from Lesson 4.3 appearing in a new role.
Two truths and a lie
All three are about rationalising.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The definition of simplified forbids a radical in a denominator only. The root of 10 over 2 is completely simplified even though a radical sits on top, and in fact rationalising typically moves the radical from the bottom to the top rather than removing it.
Prediction
Commit before reasoning.
Predict first
A calculator handles 1 over root 2 as easily as root 2 over 2. So why is the convention kept?
Correct: It makes different answers directly comparable, and hand computation easier.
\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.707 \]
Why: Both forms are exact and equal. The convention exists so that two people solving the same problem write the same answer, which matters when checking work against a key, and because dividing by a whole number is far easier by hand than dividing by an irrational one. It also makes adding fractions with radical denominators tractable, which is where it really earns its place in Chapter 5.
Section
Section 3
Concept
If a quadratic equation contains no first-power term, collect constants and divide until the equation reads x squared equals a number, then take square roots of both sides — remembering that a positive number has two square roots, so both signs must be written.
\[ 3x^2 + 5 = 41 \;\Longrightarrow\; x^2 = 12 \;\Longrightarrow\; x = \pm 2\sqrt{3} \]
If the number on the right is negative, no real solution exists, because no real number squares to a negative. Lesson 4.6 introduces the numbers that do.
Figure (svg): The parabola y equals x squared crossed by the horizontal line y equals twelve at two points
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 267-267 — Solve a quadratic equation
Picture it
The parabola y equals x squared meets any horizontal line above the origin twice.
Figure (svg): The parabola y equals x squared crossed by the horizontal line y equals twelve at two points
Squaring destroys the sign, so undoing it has to restore both possibilities. Forgetting the negative root loses half the answer, and the book flags it in an Avoid Errors note.
Worked example
Example 3, with the book's own check.
\[ \text{Solve } 3x^2 + 5 = 41. \]
Subtract to isolate the squared term
Why: Taking 5 from both sides leaves 3x squared equal to 36.
\[ 3 x ^{2} = 36 \]
Divide by the coefficient
Why: Dividing both sides by 3 gives x squared equal to 12.
\[ x ^{2} = 12 \]
Take square roots of both sides
Why: Both a positive and a negative number square to 12, so both signs are written.
\[ x = +- \sqrt{12} \]
Simplify the radical
Why: Twelve is 4 times 3, and 4 is a perfect square.
\[ x = +- 2 \sqrt{3} \]
Figure (svg): The solution to Worked example solve by taking roots shown as a ladder of expressions, one row per algebraic move
\[ x = \pm 2\sqrt{3} \]
Verify: substitute both solutions
Why: For 2 root 3: squaring gives 4 times 3, or 12, then 3 times 12 is 36, plus 5 is 41. For negative 2 root 3: squaring again gives 12, because the negative disappears, so the arithmetic is identical and the result is 41. Both check, which is the book's own verification.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 267-267
Ranking
Solving an equation with no first-power term.
Put in order
Why: Dividing before taking roots keeps the arithmetic under the radical as small as possible. The plus-or-minus goes in at the moment of the root, not afterwards, because that is the step where information about the sign is being restored.
Worked example
Guided Practice 17 to 19. The last one has a binomial inside the square.
\[ \text{Solve } 5x^2 = 80, \; z^2 - 7 = 29, \; 3(x-2)^2 = 40. \]
First: divide, then take roots
Why: Dividing by 5 gives x squared equal to 16, whose roots are 4 and negative 4.
\[ x = +- 4 \]
Second: add, then take roots
Why: Adding 7 gives z squared equal to 36, whose roots are 6 and negative 6.
\[ z = +- 6 \]
Third: divide by 3 first
Why: The equation becomes the quantity x minus 2, squared, equal to 40 over 3.
\[ (x - 2) ^{2} = \frac{40}{3} \]
Take roots and simplify
Why: The root of 40 over 3 is the root of 40 over the root of 3, which rationalises to 2 root 30 over 3.
\[ x - 2 = +- 2 \sqrt{30} / 3 \]
Isolate x
Why: Adding 2 to both sides finishes it.
\[ x = 2 + - 2 \sqrt{30} / 3 \]
Figure (svg): The solution to Worked example three more equations shown as a ladder of expressions, one row per algebraic move
\[ x = \pm 4; \quad z = \pm 6; \quad x = 2 \pm \tfrac{2\sqrt{30}}{3} \]
Verify: approximate the third answer
Why: Two root 30 over 3 is about 3.65, so the roots are about 5.65 and about negative 1.65. Substituting 5.65 gives the quantity 3.65 squared, or about 13.3, times 3, which is about 40 — as required. The two roots are equally far from 2, which is the symmetry you would expect around the vertex.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-268
Trap
\[ 3x^2 + 5 = 41 \;\Longrightarrow\; x^2 = 12 \]
Take the square root of both sides
Why: The radical symbol is applied and its output taken as the answer.
\[ x = \sqrt{12} = 2\sqrt{3} \quad \text{(only)} \qquad \text{(wrong)} \]
Negative 2 root 3 also satisfies the equation, and it has been lost. The equation is quadratic, so two solutions were to be expected.
\[ x^2 = 12 \;\Longrightarrow\; x = \pm\sqrt{12} = \pm 2\sqrt{3} \]
Write the plus-or-minus at the moment the root is taken
Why: The radical symbol denotes the principal root only, so the negative one has to be added by hand.
\[ (2\sqrt{3})^2 = 12 \qquad (-2\sqrt{3})^2 = 12 \]
A useful habit: after solving any quadratic, ask whether you have two answers, one repeated answer, or none, and be able to say which case you are in and why.
Fill the middle
Example 3, at the radical.
Fill in the blanks
x^2 = 12 \;\Longrightarrow\; x = \pm\sqrt2 = \pm\sqrt___\cdot\sqrt___ = \pm ___\sqrt___
Why: Twelve splits as 4 times 3, and the root of 4 comes out as 2, leaving 2 root 3. The plus-or-minus travels with the whole expression, so both 2 root 3 and its negative are solutions — this is one answer written compactly, not two separate calculations.
Prediction
Commit before reasoning.
Predict first
You reduce an equation to x squared equals negative 9. What are its real solutions?
Correct: There are none, because no real number squares to a negative.
\[ x^2 = -9 \;\Longrightarrow\; \text{no real solution} \quad \text{(Lesson 4.6: } x = \pm 3i \text{)} \]
Why: Every real number squares to something non-negative, so no real x can satisfy this. Graphically the parabola y equals x squared lies entirely on or above the horizontal axis, and the line y equals negative 9 never meets it. This is not a dead end, though: Lesson 4.6 introduces the imaginary unit, and in that larger number system the equation does have two solutions.
Comparison
Fill the blanks. Each method has a shape it needs.
Comparison matrix
| Question | Factoring | Taking square roots |
|---|---|---|
| What shape does it need? | factors over the integers | a square equal to a number |
| Handles x^2 + 3x - 28 = 0? | yes | no, there is a middle term |
| Handles x^2 = 12? | no, 12 is not a perfect square | yes |
| Typical answers | rational | often irrational |
Neither method is general, which is a genuine gap. Lesson 4.7 forces any quadratic into the shape the right column needs, and Lesson 4.8 turns that into a formula.
Section
Section 4
Concept
The method does not care what is inside the square. If the equation reduces to a bracket squared equal to a number, taking roots gives that whole bracket equal to plus or minus the root, and one more step isolates the variable.
\[ \tfrac{1}{7}(z-2)^2 = 6 \;\Longrightarrow\; (z-2)^2 = 42 \;\Longrightarrow\; z = 2 \pm \sqrt{42} \]
This is the shape Lesson 4.7 will manufacture deliberately for every quadratic, which is why it is worth being fluent with it now, while the bracket is handed to you.
Figure (svg): Two columns comparing solving by factoring with solving by taking square roots
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-268 — Example 4
Picture it
What each method needs, and where each fails.
Figure (svg): Two columns comparing solving by factoring with solving by taking square roots
Recognising the shape of an equation before choosing a method saves more time than any technique in the chapter.
Worked example
Example 4. Clear the fraction, then take roots.
\[ \text{Solve } \tfrac{1}{7}(z - 2)^2 = 6. \]
Multiply both sides by 7
Why: The fraction clears, leaving the bracket squared equal to 42.
\[ (z - 2) ^{2} = 42 \]
Take square roots of both sides
Why: The whole bracket equals plus or minus the root of 42.
\[ z - 2 = +- \sqrt{42} \]
Check whether the radical simplifies
Why: Forty-two is 2 times 3 times 7, with no repeated factor, so it has no perfect-square factor and cannot be simplified.
\[ \sqrt{42}\text{ stays as it is} \]
Isolate the variable
Why: Adding 2 to both sides gives the two solutions.
\[ z = 2 + - \sqrt{42} \]
Figure (svg): The solution to Worked example a binomial inside the square shown as a ladder of expressions, one row per algebraic move
\[ z = 2 \pm \sqrt{42} \]
Verify: substitute the positive solution
Why: If z is 2 plus root 42, then z minus 2 is root 42, whose square is 42, and a seventh of 42 is 6 — the right side exactly. The negative solution works identically because the squaring erases the sign, which is why the two solutions are symmetric about 2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-268
Sorting
Look for a middle term, and for whether it factors.
Sort into buckets
Sort each equation by the better method.
Two seconds of looking at the shape decides the method. Choosing wrongly costs several minutes and usually ends in an unsolvable step.
Worked example
Guided Practice 19 again, this time with the rationalising shown in full.
\[ \text{Solve } 3(x - 2)^2 = 40. \]
Divide both sides by 3
Why: The bracket squared equals 40 over 3.
\[ (x - 2) ^{2} = \frac{40}{3} \]
Take roots of both sides
Why: The bracket equals plus or minus the root of 40 over 3.
\[ x - 2 = +- \sqrt{\frac{40}{3}} \]
Split with the quotient property and simplify the top
Why: Forty is 4 times 10, so the numerator becomes 2 root 10, over the root of 3.
\[ +- 2 \sqrt{10} / \sqrt{3} \]
Rationalise the denominator
Why: Multiplying top and bottom by the root of 3 gives 2 root 30 over 3.
\[ +- 2 \sqrt{30} / 3 \]
Isolate x
Why: Adding 2 gives both solutions.
\[ x = 2 + - 2 \sqrt{30} / 3 \]
Figure (svg): The solution to Worked example a coefficient outside the bracket shown as a ladder of expressions, one row per algebraic move
\[ x = 2 \pm \frac{2\sqrt{30}}{3} \]
Verify: approximate both sides
Why: Two root 30 over 3 is about 3.651, so x is about 5.651 or about negative 1.651. Squaring 3.651 gives about 13.33, and 3 times that is about 40. The check also shows why rationalising matters practically: dividing by 3 is easy, dividing by root 3 is not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-268
Error analysis
A student solves a squared binomial by taking roots term by term.
Annotate
On: \( (z - 2)^2 = 42 \;\Longrightarrow\; z - 2 = \sqrt{42} \;\Longrightarrow\; z = \sqrt{42} - 2 \)
Isolating the variable after taking roots is an ordinary linear step, so treat it as one: whatever is done to one side is done to the other.
Fill the middle
Example 4, at the last step.
Fill in the blanks
z - 2 = \pm\sqrt2 \;\Longrightarrow\; z = ___ \pm \sqrt___
Why: Adding 2 to both sides moves it to the right, giving 2 plus or minus root 42. The plus-or-minus stays attached to the radical, not to the 2, so the two solutions are symmetric about 2 — which makes sense, since 2 is the value that makes the bracket vanish.
Prediction
Commit before reasoning.
Predict first
The equation reads a times the quantity x minus h, squared, equals a number. What does that remind you of?
Correct: Vertex form from Lesson 4.2, set equal to a value.
\[ 3(x-2)^2 = 40 \;\Longleftrightarrow\; y = 3(x-2)^2 \text{ meets } y = 40 \]
Why: Three times the quantity x minus 2, squared, equals 40 is exactly the question where does the parabola y equals 3 times the quantity x minus 2, squared, reach the height 40 — and the two solutions are the two x values at that height, symmetric about the axis x equals 2. Seeing the vertex form inside these equations is what makes Lesson 4.7's completing the square feel inevitable rather than arbitrary.
Comparison
Fill the blanks. The difference is what sits inside the square.
Comparison matrix
| Step | 3x^2 + 5 = 41 | 3(x - 2)^2 = 40 |
|---|---|---|
| Isolate the square | x^2 = 12 | (x - 2)^2 = 40/3 |
| Take roots | x = +- 2 sqrt(3) | x - 2 = +- sqrt(40/3) |
| Rationalise? | not needed | yes, giving 2 sqrt(30)/3 |
| Extra step at the end | none | add 2 to isolate x |
The right column has two extra steps and no new ideas. That is the general shape of this lesson: the method is short, and the presentation of the answer is where the work is.
Section
Section 5
Concept
An object dropped from rest has height, in feet, given by negative 16t squared plus its initial height, with t in seconds. Setting the height to zero and solving asks when it reaches the ground, and because there is no first-power term the equation is solved by taking roots.
\[ h = -16t^2 + h_0 \]
The model assumes air resistance is negligible and that the object is dropped rather than thrown. The 16 encodes Earth's gravity in feet per second squared, so a different planet needs a different number.
Figure (svg): The height of a dropped object against time, from fifty feet down to the ground
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 268-269 — Modeling dropped objects
Picture it
Example 5: an egg container dropped from 50 feet.
Figure (svg): The height of a dropped object against time, from fifty feet down to the ground
Only the right half of the parabola is drawn, because negative times have no meaning here. The negative root is discarded for exactly that reason.
Worked example
Example 5. Substitute, isolate, take roots, reject.
\[ \text{An egg container is dropped from } 50 \text{ feet. How long until it lands?} \]
Write the model and substitute
Why: The height is zero at the ground and the initial height is 50.
\[ 0 = -16 t ^{2} + 50 \]
Isolate the squared term
Why: Subtracting 50 and dividing by negative 16 gives t squared equal to 50 over 16.
\[ t ^{2} = \frac{50}{16} \]
Take square roots
Why: Both signs are written, as always.
\[ t = +- \sqrt{\frac{50}{16}} \]
Evaluate and reject
Why: The root of 50 over 16 is about 1.8, and a negative time is meaningless here.
\[ t =\text{ about } 1.8 \]
Figure (svg): The solution to Worked example the falling container shown as a ladder of expressions, one row per algebraic move
\[ t \approx 1.8 \text{ seconds} \]
Verify: substitute the time back
Why: At t equal to 1.8, t squared is 3.24, and 16 times that is about 51.8, so the height is 50 minus 51.8, which is about negative 1.8 feet — slightly below ground because 1.8 is a rounded value. Using the exact root of 50 over 16 gives exactly zero, which is the right check to trust.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 269-269
Fill the middle
Example 5, at the substitution.
Fill in the blanks
h = -16t^2 + h_0, \; h = 0, \; h_0 = 50 \;\Longrightarrow\; t^2 = \frac16___}
Why: Subtracting 50 gives negative 50 equal to negative 16t squared, and dividing both sides by negative 16 makes both signs positive, leaving 50 over 16. The two negatives cancelling is what keeps the right side positive, which it has to be for a real solution to exist.
Worked example
Guided Practice 20. Notice what halving the height does to the time.
\[ \text{Repeat for a drop from } 30 \text{ feet.} \]
Substitute the new initial height
Why: The model becomes zero equals negative 16t squared plus 30.
\[ 0 = -16 t ^{2} + 30 \]
Isolate and take roots
Why: T squared is 30 over 16, which is 1.875.
\[ t = +- \sqrt{1.875} \]
Evaluate and reject the negative
Why: The positive root is about 1.4 seconds.
\[ t =\text{ about } 1.4 \]
Compare with the 50 foot drop
Why: The height fell by 40 percent but the time fell by only about 22 percent, because time depends on the square root of the height.
Figure (svg): The solution to Worked example a shorter drop shown as a ladder of expressions, one row per algebraic move
\[ t \approx 1.4 \text{ seconds} \]
Verify: check the ratio
Why: The ratio of the two times should be the square root of the ratio of the heights: 30 over 50 is 0.6, and the root of 0.6 is about 0.775. And 1.4 divided by 1.8 is about 0.78, which agrees. Doubling a drop height therefore multiplies the fall time by only about 1.41, not by 2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 269-269
Error analysis
A student solves the drop from 50 feet and reports two answers.
Annotate
On: \( t = \pm\sqrt{\tfrac{50}{16}} \approx \pm 1.8, \text{ so the container lands at } t = 1.8 \text{ and at } t = -1.8 \)
The same rejection as the field and quilt problems of the last two lessons. The equation describes more than the situation does, and the extra solution is discarded on physical grounds, not mathematical ones.
Ranking
Answering a dropped-object question.
Put in order
Why: Stating what the letters mean is what makes the rejection in step four justifiable rather than a habit. Note that the height being substituted is not always zero: asking when the container passes 20 feet substitutes 20 there instead, and the method is otherwise identical.
Prediction
Commit before reasoning.
Predict first
The model contains the initial height but not the object's mass. What does that say?
Correct: In this model, all dropped objects fall at the same rate regardless of mass.
\[ h = -16t^2 + h_0: \; \text{no mass appears anywhere} \]
Why: The absence of mass is a genuine physical claim, not an oversight, and it is one of the most famous results in mechanics. It holds because the model neglects air resistance; a feather and a hammer really do land together in a vacuum. Where air resistance matters, the model stops applying — which the textbook states explicitly rather than leaving to be discovered.
Comparison
Fill the blanks. Time does not scale with height.
Comparison matrix
| Quantity | From 50 feet | From 30 feet |
|---|---|---|
| Equation | 0 = -16t^2 + 50 | 0 = -16t^2 + 30 |
| t^2 equals | 50/16 = 3.125 | 30/16 = 1.875 |
| Fall time | about 1.8 seconds | about 1.4 seconds |
| Ratio of times | 1.4/1.8, about 0.78 | the square root of 30/50, about 0.77 |
Times scale with the square root of the height, which is why a drop from four times the height takes only twice as long.
Comparison
Fill the blanks. Each needs a particular shape.
Comparison matrix
| Method | Needs | Fails on |
|---|---|---|
| Factoring, a = 1 | integer factors of c summing to b | x^2 = 12 |
| Factoring, a not 1 | integer factors of a and c that fit | x^2 - 2 = 0 |
| Taking square roots | a square equal to a number | x^2 + 3x - 28 = 0 |
| All three | give exact answers | none is fully general |
The gap is real and the chapter closes it deliberately: Lesson 4.7 turns any quadratic into the third row's shape, and Lesson 4.8 packages the result as a formula that always works.
Pattern
One routine for any equation with no first-power term.
If the number on the right comes out negative, stop: there is no real solution, and Lesson 4.6 is where that case is answered.
OpenStax Algebra and Trigonometry 2e, §2.5 Quadratic Equations §2.5
Check
Simplifying. Find the largest square factor.
Check your understanding
Simplify sqrt(8) x sqrt(28).
Answer: A
Why: The product is sqrt(224), and 224 is 16 times 14, so the 16 comes out as 4, leaving 4 sqrt(14).
Check
Solving by taking roots. Both signs.
Check your understanding
Solve 3x^2 + 5 = 41.
Answer: A
Why: Subtracting 5 and dividing by 3 gives x^2 = 12, so x = +- sqrt(12) = +- 2 sqrt(3).
Check
A binomial inside the square.
Check your understanding
Solve (1/7)(z - 2)^2 = 6.
Answer: A
Why: Multiplying by 7 gives (z - 2)^2 = 42, so z - 2 = +- sqrt(42) and z = 2 +- sqrt(42).
Real world
A square photograph is to be enlarged so that its area is exactly 2.5 times the original. The original is 8 inches on a side.
Discussion prompt
Find the side length of the enlargement, exactly and to the nearest tenth of an inch. Then say what percentage each side grew by, and why that is not 150 percent.
Hint: Let s be the new side. The new area is s squared and the old area is 64.
Answer:
\[ s^2 = 2.5(64) = 160 \;\Longrightarrow\; s = \pm\sqrt{160} = \pm 4\sqrt{10} \]
The negative root is rejected, so the side is 4 root 10, or about 12.6 inches. Each side grew by about 58 percent, not 150 percent.
The reason is that area scales with the square of the side. Multiplying the area by 2.5 multiplies the side by the square root of 2.5, which is about 1.58. This is the same square-root scaling as the two drop heights, and it is one of the most useful facts in the lesson: whenever a quantity depends on a square, its inverse depends on a square root, and proportional reasoning has to be done through that root.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the root of 5 over the root of 2 equal to the root of 10 over 2?
Correct: Yes — they are the same number written two ways.
\[ \frac{\sqrt{5}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{10}}{2} \approx 1.5811 \]
Why: Multiplying by root 2 over root 2 is multiplying by 1, and multiplying by 1 cannot change a value. Both expressions are about 1.5811. Rationalising is a change of form only, and this is worth being certain about, because the whole practice of preferring one form over the other rests on the two being genuinely equal.
Explain it
They can solve equations by factoring and think that is all there is.
Discussion prompt
In four sentences or fewer, explain when taking square roots beats factoring, and why the answer always has two parts.
Hint: Point at whether there is a first-power term.
Answer:
If the equation has no plain x term — just a square and some constants — you can get it to the form something squared equals a number and undo the square directly. Factoring would need that number to be a perfect square, and usually it is not, so this method reaches answers factoring cannot.
There are two answers because squaring throws away the sign: both 3 and negative 3 square to 9. So when you undo a square you have to put both possibilities back, which is what the plus-or-minus sign means.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For square factors, run down the list 4, 9, 16, 25, 36, 49 and test each against the radicand rather than guessing. For conjugates, write the denominator down and change only the sign between its two terms. For the plus-or-minus, write it at the same moment you write the radical sign, never afterwards. For isolating, treat it as an ordinary linear step and do the same thing to both sides. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the equation 3 times the quantity x minus 2, squared, equals 40 and build the page around it. Top left: solve it completely, showing the division, the root with its plus-or-minus, the simplification of the radical and the rationalisation, one line per move. Top right: write the two clauses of the definition of simplified and tick off where in your solution each one was satisfied. Bottom left: sketch y equals 3 times the quantity x minus 2, squared, and the horizontal line y equals 40, marking both intersection points and the axis of symmetry between them. Bottom right: write the same equation with the 40 replaced by 0 and then by negative 40, and say how many real solutions each version has and why. In a margin, write the conjugate of 7 minus root 5 and the rational number their product gives.
If your bottom-right panel says the version with negative 40 has two solutions, check it against your sketch: the line y equals negative 40 lies entirely below the parabola.
Recap
Five things, and the middle one is the method the rest support.
| If you see | Then |
|---|---|
| A radicand with a square factor | Split it off with the product property |
| A radical in a denominator | Multiply by it, or by the conjugate |
| A square equal to a number | Take roots of both sides, with plus-or-minus |
| A square equal to a negative | No real solution; see Lesson 4.6 |
| A bracket squared | The root undoes the whole bracket |
| A negative time or length | Reject it |
The one case this lesson cannot answer is a square equal to a negative number. Lesson 4.6 builds the numbers that answer it, and with them every quadratic equation gets solutions.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.5 Solve Quadratic Equations by Finding Square Roots §4.5, pp. 266-269 — everything on these slides traces back here
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