UNIT 3Quadratic functionsLesson 2 of 31 Solving Quadratics by Factoring2 Solving Quadratics by Square Roots3 Graphing Quadratic Functions

Solving Quadratics by Square Roots

A1.7Algebra 1Functions20 practice questions

When an equation is a square equal to a number, take the square root of both sides. Remember the ± sign.

What you'll be able to do

Solve x² = k and (x − h)² = k.
Get the square alone before you take a square root.
Tell how many real solutions an equation has.
Give exact answers, and round when a question asks.

The big idea

Two answers. x² = 49 means x = 7 or x = −7, because both square to 49. We write x = ±7.

Get the square alone first. In 3x² − 12 = 0, add 12 and divide by 3: x² = 4. Then x = ±2.

A square with a shift. (x − 3)² = 25 means x − 3 = ±5. So x = 8 or x = −2.

Exact form. x² = 20 gives x = ±√20 = ±2√5. Round only when the question asks.

How many solutions? If the square equals a positive number, there are 2. If it equals 0, there is 1. If it equals a negative number, there are none, because no real number squared is negative.

Check the story. A side length or a time can't be negative. Throw out the negative answer.

Words you need

Square root: A number that gives the first number when you multiply it by itself. √49 = 7.
Plus or minus: The sign ±. It means two answers: x = ±7 is x = 7 or x = −7.
Quadratic equation: An equation with an x² term and no higher power.
Solution: A value of x that makes the equation true.
Real number: Any number on the number line.
Simplest radical form: No perfect square is left under the root sign, like 2√5 instead of √20.
Radical: A root sign and what is under it, like √5.
Perfect square: A number or term that is something times itself, like 49 or (x + 5)².
Perfect square trinomial: A trinomial that equals a binomial squared, like x² + 6x + 9 = (x + 3)².
Hypotenuse: The longest side of a right triangle. It is across from the right angle.
Factor: A part you multiply. To factor means to write something as a product.
Product: The answer when you multiply.
Coefficient: The number in front of a letter. In 3x², it is 3.
Constant: A term with no letter, like −12.
Integer: A whole number or its negative, like −3, 0 or 7.
Exponent: The small raised number. In t², it is 2.
Power: A base with an exponent, like x².
Base: The bottom of a shape, like the base of a wall. In a power like 3², the base is 3.
Variable: A letter that stands for a number, like x or t.
Expression: Numbers and letters joined by +, −, × or ÷. It has no = sign.
Simplify: Write it in a shorter form with the same value.
Denominator: The bottom number of a fraction.
Imaginary unit: The number i, where i² = −1. Then √(−9) = 3i.

How many real solutions?

Equation	Real solutions	Why
x² = 9	2: x = 3 and x = −3	9 is positive
x² = 0	1: x = 0	Only 0 squares to 0
x² = −9	None	No real number squared is negative

Equation

Real solutions

Why

x² = 9

2: x = 3 and x = −3

9 is positive

x² = 0

1: x = 0

Only 0 squares to 0

x² = −9

None

No real number squared is negative

Worked examples

Example 1

Solve x² = 50. Give the answers in simplest radical form.

Take the square root of both sides: x = ±√50.
50 = 25 · 2, so √50 = 5√2.
x = 5√2 or x = −5√2.

Answer: x = 5√2 or x = −5√2

Example 2

Solve 2x² + 3 = 75.

Get x² alone. Subtract 3: 2x² = 72.
Divide by 2: x² = 36.
Take the square root: x = 6 or x = −6.

Answer: x = 6 or x = −6

Example 3

Solve (x + 1)² = 18. Give the answers in simplest radical form.

Take the square root of both sides: x + 1 = ±√18.
√18 = √(9 · 2) = 3√2. So x + 1 = ±3√2.
Subtract 1: x = −1 + 3√2 or x = −1 − 3√2.

Answer: x = −1 + 3√2 or x = −1 − 3√2

Example 4

A stone falls from a cliff 64 feet high. The distance it falls in t seconds is d = 16t² feet. How long does it take to hit the ground?

Set d = 64: 16t² = 64.
Divide by 16: t² = 4. So t = 2 or t = −2.
Time can't be negative. The stone hits the ground after 2 seconds.

Answer: 2 seconds

Common mistakes

Forgetting the ± sign. x² = 36 has two solutions, 6 and −6.

Taking the root too soon. In 2x² + 5 = 37, get x² alone first: x² = 16, so x = ±4.

Splitting a root over a sum. √(x² + 36) is not x + 6. Solve x² + 36 = 100 by subtracting 36 first: x² = 64, so x = ±8.

Keeping a negative length. A side of −3√5 feet is not real. Throw it out.

Practice: 20 questions

11 of 20 are word problems. For a word problem, sketch it first and mark what you know.

Tier 1 Foundations Use the rule directly

Find all real solutions to the equation x² = 72. Give your answers in simplest radical form.
Priya drops a geology marker from a bridge 196 feet above a canyon floor. The distance d, in feet, that an object falls after t seconds is given by the formula d = 16t². Find how many seconds the marker takes to fall 196 feet, and explain why one solution of the equation is thrown out. Give your answer as a decimal or fraction in seconds. Word problem
A square solar panel on a satellite has an area of 45 square feet. The area A of a square with side length s is given by the formula A = s². Find the side length of the panel, and state why the negative solution is rejected. Give both the exact answer in simplest radical form and the answer rounded to the nearest tenth of a foot. Word problem
Kenji solves the equation 3x² − 48 = 0 by adding 48 to both sides to get 3x² = 48, dividing by 3 to get x² = 16, and writing x = 4 as the only answer. Explain what is wrong with Kenji's final step, and give all real solutions to the equation. Give your answers as integers.
Without finding the solutions, compare three equations. Equation A is x² − 19 = 0. Equation B is 4x² + 25 = 0. Equation C is 9x² = 0. Which equation has the most real solutions, and which equation has the fewest real solutions? State the number of real solutions for each equation.

Tier 2 Intermediate One more step to think through

Zahra is designing a circular fountain whose basin must have an area of 108π square feet. The area A of a circle with radius r is given by the formula A = πr². Find the exact radius of the fountain in simplest radical form, and explain why one algebraic solution is thrown out. Give your answer in feet. Word problem
Devon has a square wooden deck with side length x feet. He adds 4 feet to both the length and the width to create a larger square deck with an area of 169 square feet, modeled by the equation (x + 4)² = 169. Find all algebraic solutions to (x + 4)² = 169, and state the original side length x of the deck in feet. Word problem
An 11-foot ladder leans against a vertical wall with the bottom of the ladder resting on level ground 5 feet from the base of the wall. By the Pythagorean theorem, a² + b² = c², where a and b are the legs of a right triangle and c is the hypotenuse. Find how high up the wall the top of the ladder reaches. Give both the exact height in simplest radical form and the height rounded to the nearest tenth of a foot. Word problem
Luciana wants to solve x² + 16 = 25. She takes the square root of each term to write x + 4 = ±5, and gets x = 1 and x = −9. Explain why Luciana's first step is invalid, and find all correct real solutions to the equation. Give your answers as integers.
Ibrahim and Freya both solve 4x² − 49 = 0. Ibrahim factors the left side as a difference of squares, while Freya isolates x² and takes square roots. First, state whether both methods give the same solutions for 4x² − 49 = 0. Second, compare the two methods on the equation 4x² − 20 = 0, and explain which method still works with integer factoring versus radicals. Give all real solutions to both equations in simplest form.

Tier 3 Advanced Several steps - find the hidden structure

A square courtyard has sides of length (2x − 3) meters and an area of 50 square meters. Find all real solutions to the area equation (2x − 3)² = 50 in simplest radical form, and identify which solution gives a positive side length for the courtyard. Word problem
Astrid places a 5-inch-wide wooden border along two adjacent sides of a square mirror of side length x inches, forming a larger square whose total area in square inches is x² + 10x + 25. This total area is 144 square inches. Solve x² + 10x + 25 = 144, and state the side length x of the mirror in inches. Word problem
A weather drone hovers 320 feet above a flat desert testing ground and drops a sensor package. The distance d, in feet, that the sensor falls in t seconds is given by the formula d = 16t². How many seconds does the sensor take to fall the first 144 feet, and how many additional seconds does it take to hit the ground 320 feet below the drone? Explain why negative time solutions are thrown out, and round the additional time to the nearest hundredth of a second. Word problem
Emiliano has three identical square tiles with sides of x − 2 inches. Their total area is 36 square inches, so 3(x − 2)² = 36. He takes the square root of both sides first to write 3(x − 2) = ±6, and gets x = 4 and x = 0. Explain the error in his first step. Find both solutions in simplest radical form, and give the real side length of a tile. Word problem
Compare Equation A, which is x² − 14x + 49 = 36, and Equation B, which is 2(x − 3)² − 7 = 91. Which equation has the larger greatest solution, and which equation has the smaller least solution? Give all solutions to both equations as integers.

Tier 4 Challenge Stretch problems, some above grade level

Yara designs a rectangular kite sail whose length is twice its width w. A straight carbon-fiber rod runs diagonally from corner to corner across the sail and measures 15 feet long. By the Pythagorean theorem, the sum of the squares of the width and the length equals the square of the diagonal. Find the exact width and length of the sail in simplest radical form, in feet. Word problem
A circular flower bed has an area of 49π square feet. A stone walkway of uniform width w feet is built all the way around the outside of the flower bed, so the flower bed and walkway together form a larger circle with a total area of 128π square feet. With the circle area formula A = πr², find the exact width w of the walkway in simplest radical form, and round w to the nearest tenth of a foot. Word problem
Desmond solves the equation (2x − 5)² = (x + 1)² by taking the square root of both sides, writing 2x − 5 = x + 1, and concluding that x = 6 is the only solution. Explain why Desmond lost a solution, and find all real solutions to the equation. Give your answers as integers or fractions in simplest form.
Let a and c be positive numbers. Consider Equation A, which is a(x − 4)² + c = 0, and Equation B, which is a(x − 4)² − c = 0. First, determine which of the two equations has real solutions. Second, for the equation that has real solutions, explain how multiplying c by 4 changes the distance between its two solutions.
In higher algebra, the imaginary number i is defined by the rule i² = −1. Because i² = −1, the square root of any negative number can be written in terms of i, such as √(−9) = 3i. Solve the equation x² − 6x + 9 = −27 over the complex numbers, and write both solutions in the form a + bi with simplest radicals. Above grade level

Answers

Tier 1 Foundations

x = 6√2 and x = −6√2. Take the square root of both sides of x² = 72 to get x = ±√72. Factor 72 as 36 · 2 so that √72 = √36 · √2 = 6√2. The two solutions are x = 6√2 and x = −6√2.
3.5 seconds, or 7/2 seconds; t = −3.5 is thrown out because time after dropping cannot be negative. Substitute d = 196 into the formula to get 16t² = 196, and divide both sides by 16 to get t² = 196/16 = 49/4. Take the square root of both sides to get t = ±7/2, which is t = 3.5 or t = −3.5. Because time cannot be negative, t = −3.5 is thrown out, leaving 3.5 seconds.
Exact: 3√5 feet; rounded: 6.7 feet; −3√5 is rejected because the side length of a square cannot be negative. Substitute A = 45 into A = s² to get s² = 45, so s = ±√45 = ±3√5. Reject −3√5 because a physical side length cannot be negative. Since 3√5 is approximately 6.708, it rounds to 6.7 feet.
Kenji forgot the negative square root when solving x² = 16; the solutions are x = 4 and x = −4. Both a positive number and its negative opposite square to the same positive value. Because (−4)² = 16 just as 4² = 16, taking the square root of both sides of x² = 16 gives x = ±4. The complete solutions are x = 4 and x = −4.
Equation A has the most real solutions and Equation B has the fewest; Equation A has 2 real solutions, Equation B has 0 real solutions, and Equation C has 1 real solution. Isolating x² in Equation A gives x² = 19, which is greater than 0 and therefore has 2 real solutions. Isolating x² in Equation B gives x² = −25/4, which is less than 0 and has 0 real solutions because a real square cannot be negative. Isolating x² in Equation C gives x² = 0, which has 1 real solution.

Tier 2 Intermediate

6√3 feet; r = −6√3 is thrown out because the radius of a circle cannot be negative. Substitute A = 108π into A = πr² to get πr² = 108π. Divide both sides by π to isolate r² = 108, then take the square root of both sides to get r = ±√108 = ±6√3. Throw out −6√3 because a radius must be positive, leaving 6√3 feet.
The algebraic solutions are x = 9 and x = −17; the original side length of the deck is 9 feet. Take the square root of both sides of (x + 4)² = 169 to get x + 4 = ±13. Subtract 4 from both sides to get x = 13 − 4 = 9 or x = −13 − 4 = −17. Because a deck's side length cannot be negative, the original side length is 9 feet.
Exact: 4√6 feet; rounded: 9.8 feet. Substitute a = 5 and c = 11 into a² + b² = c² to get 5² + b² = 11², or 25 + b² = 121. Subtract 25 from both sides to get b² = 96, so b = √96 = 4√6 after rejecting the negative root. Since 4√6 is approximately 9.798, it rounds to 9.8 feet.
You cannot take the square root of each term in a sum separately because √(x² + 16) does not equal x + 4; the correct solutions are x = 3 and x = −3. Square roots do not distribute across addition, so x² must be isolated before taking the square root. Subtract 16 from both sides of x² + 16 = 25 to get x² = 9. Taking the square root of both sides gives x = 3 and x = −3.
Both methods give the same solutions for 4x² − 49 = 0, which are x = 7/2 and x = −7/2; for 4x² − 20 = 0, integer factoring stops after 4(x² − 5) = 0 because 5 is not a perfect square, whereas Freya's square root method gives x = √5 and x = −√5. For 4x² − 49 = 0, factoring (2x − 7)(2x + 7) = 0 and solving x² = 49/4 both yield x = ±7/2. For 4x² − 20 = 0, factoring out 4 leaves 4(x² − 5) = 0, which cannot be factored further with integers. Isolating x² gives x² = 5, so taking square roots directly gives x = √5 and x = −√5.

Tier 3 Advanced

All solutions: x = (3 + 5√2)/2 and x = (3 − 5√2)/2; only x = (3 + 5√2)/2 gives a positive side length. Take the square root of both sides of (2x − 3)² = 50 to get 2x − 3 = ±√50 = ±5√2. Add 3 and divide by 2 to get x = (3 + 5√2)/2 and x = (3 − 5√2)/2. Because the side length is 2x − 3, choosing x = (3 − 5√2)/2 makes the side length −5√2 meters, which is negative, so only x = (3 + 5√2)/2 is valid.
7 inches. Recognize that x² + 10x + 25 is a perfect square trinomial and rewrite the equation as (x + 5)² = 144. Take the square root of both sides to get x + 5 = ±12, which gives x = 7 or x = −17. Since the mirror's side length must be positive, x = 7 inches.
3 seconds for the first 144 feet, and 1.47 additional seconds to hit the ground; negative times are thrown out because time after the drop cannot be negative. For the first 144 feet, solve 16t² = 144 to get t² = 9, so t = 3 seconds after rejecting t = −3. For the full 320-foot drop, solve 16t² = 320 to get t² = 20, so t = √20 = 2√5 ≈ 4.472 seconds after rejecting t = −2√5. Subtract 3 from 4.472 to get 1.472, which rounds to 1.47 additional seconds.
Emiliano took the square root of (x − 2)² and 36 without dividing by 3 or taking the square root of the coefficient 3; the solutions are x = 2 + 2√3 and x = 2 − 2√3, and a tile's side is x − 2 = 2√3 inches, because x = 2 − 2√3 makes the side negative. Before taking the square root of both sides, the squared binomial (x − 2)² must be isolated. Divide both sides of 3(x − 2)² = 36 by 3 to get (x − 2)² = 12. Take the square root of both sides to get x − 2 = ±√12 = ±2√3, so x = 2 + 2√3 and x = 2 − 2√3. The side x − 2 must be positive, so it is 2√3 inches.
Equation A has the larger greatest solution, and Equation B has the smaller least solution; Equation A's solutions are 13 and 1, while Equation B's solutions are 10 and −4. Rewrite Equation A as (x − 7)² = 36, so x − 7 = ±6, giving x = 13 and x = 1. For Equation B, add 7 and divide by 2 to get (x − 3)² = 49, so x − 3 = ±7, giving x = 10 and x = −4. Comparing the solutions shows 13 is greater than 10, while −4 is less than 1.

Tier 4 Challenge

Width: 3√5 feet; length: 6√5 feet. The width is w and the length is 2w, so w² + (2w)² = 15². Squaring 2w gives 4w², so w² + 4w² = 225, or 5w² = 225. Divide by 5 to get w² = 45, and take the positive square root to get a width of w = √45 = 3√5 feet and a length of 2(3√5) = 6√5 feet.
Exact: 8√2 − 7 feet; rounded: 4.3 feet. For the inner circle, πr² = 49π gives r² = 49, so the inner radius is 7 feet. For the outer circle, πR² = 128π gives R² = 128, so the outer radius is R = √128 = 8√2 feet. Subtract the inner radius from the outer radius to get the walkway width w = 8√2 − 7 feet, which is approximately 11.314 − 7 = 4.314 and rounds to 4.3 feet.
Desmond forgot the ± when taking the square root of both sides, missing the case where 2x − 5 = −(x + 1); the solutions are x = 6 and x = 4/3. Whenever two squares are equal, A² = B², taking the square root of both sides gives A = B or A = −B. Setting 2x − 5 = x + 1 gives x = 6. Setting 2x − 5 = −(x + 1) gives 2x − 5 = −x − 1, so 3x = 4 and x = 4/3.
Only Equation B has real solutions; multiplying c by 4 doubles the distance between Equation B's two solutions. Isolating (x − 4)² gives (x − 4)² = −c/a for Equation A and (x − 4)² = c/a for Equation B. Since a and c are both positive, −c/a is negative and c/a is positive, so only Equation B has real solutions: x = 4 + √(c/a) and x = 4 − √(c/a). The distance between these two solutions is 2√(c/a), and replacing c with 4c multiplies √(c/a) by √4 = 2, which doubles the distance between the solutions.
x = 3 + 3i√3 and x = 3 − 3i√3. Rewrite the perfect square trinomial x² − 6x + 9 as (x − 3)², giving (x − 3)² = −27. Take the square root of both sides to get x − 3 = ±√(−27). Since √(−27) = √27 · √(−1) = 3i√3, add 3 to both sides to get x = 3 + 3i√3 and x = 3 − 3i√3.

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