Algebra 2 Quiz: Complete The Square To Find Solutions

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Complete The Square To Find SolutionsQuestion 1 of 20

Solve the quadratic equation by completing the square: $x^2 + 8x + 7 = 0.$ Write your work in the form $(x - p)^2 = q$ before solving.

(x+4)^2=9

, so

x=-4 \pm 3

(i.e.,

x=-1

x=-7

(x+8)^2=57

, so

x=-8 \pm \sqrt{57}

(x+4)^2=7

, so

x=-4 \pm \sqrt{7}

(x-4)^2=9

, so

x=4 \pm 3

(i.e.,

x=1

x=7

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Algebra 2 Quiz

Algebra 2 Quiz: Complete The Square To Find Solutions

Practice Complete The Square To Find Solutions in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complete The Square To Find Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve the quadratic equation by completing the square: $x^2 + 8x + 7 = 0.$ Write your work in the form $(x - p)^2 = q$ before solving.

$(x+4)^2=9$ , so $x=-4 \pm 3$ (i.e., $x=-1$ or $x=-7$ ). (correct answer)
$(x+8)^2=57$ , so $x=-8 \pm \sqrt{57}$ .
$(x+4)^2=7$ , so $x=-4 \pm \sqrt{7}$ .
$(x-4)^2=9$ , so $x=4 \pm 3$ (i.e., $x=1$ or $x=7$ ).

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If

q

is positive, you get two real solutions; if

q = 0

, one solution; if

q

negative, two complex solutions

x = p \pm i \sqrt{|q|}

. The value of

q

immediately reveals the nature of solutions. To solve

x^2 + 8x + 7 = 0

by completing the square: (1) Move constant to right:

x^2 + 8x = -7

. (2) Take half of 8 to get 4, square it to get 16, add to both sides:

x^2 + 8x + 16 = -7 + 16 = 9

. (3) Factor left side as perfect square:

(x + 4)^2 = 9

. (4) Take square roots:

x + 4 = \pm 3

. (5) Solve:

x = -4 \pm 3

, so

x = -1

x = -7

. The systematic process works every time! For deriving the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants, following the same steps but keeping everything symbolic. Choice A correctly completes the square to get the proper form and solves to find both solutions accurately. Choice B calculates

(b/2)^2

incorrectly: half of 8 is 4, and 4 squared is 16, not leading to

(x+8)^2=57

. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If

a \neq 1

, divide everything by

a

first to get

x^2

coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for

x

. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants: divide by

a

to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 2

When deriving the quadratic formula from $ax^2 + bx + c = 0$ using completing the square, the intermediate step $a(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a}$ is reached. What is the next step that leads directly to the quadratic formula?

Expand the left side, then rearrange to solve for $x$
Take the square root of both sides, then subtract $\frac{b}{2a}$ from both sides
Divide both sides by $a$ , then take the square root of both sides (correct answer)
Factor the right side, then divide both sides by $4a$

Explanation: When you're completing the square to derive the quadratic formula, you need to recognize what form you're working toward and what operations will get you there most efficiently. Starting from

a(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a}

, the most direct path is to divide both sides by

a

first, giving you

(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}

. Then take the square root of both sides:

x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}

. Finally, subtract

\frac{b}{2a}

to get

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

— the quadratic formula. This matches choice C. Choice A is inefficient and backward — expanding the perfect square would undo the completing-the-square work you just did. Choice B has the right operations but in the wrong order; if you take the square root before dividing by

a

, you get

x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2\sqrt{a}}

, which creates an awkward square root in the denominator that doesn't lead cleanly to the standard form. Choice D suggests factoring

b^2 - 4ac

, but this expression (the discriminant) generally can't be factored in a useful way, and dividing by

4a

doesn't advance you toward the solution. Remember: when you have a perfect square equal to something, always isolate the squared term first (by dividing out coefficients), then take the square root. This order prevents messy expressions and keeps you on the most direct path to your goal.

Question 3

Consider the quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$ . When this equation is transformed by completing the square to the form $a(x - p)^2 = q$ , which expression correctly represents $q$ in terms of $a$ , $b$ , and $c$ ?

$q = \frac{b^2 - 4ac}{4a}$ (correct answer)
$q = \frac{4ac - b^2}{4a}$
$q = \frac{b^2 - 4ac}{4}$
$q = \frac{4ac - b^2}{4}$

Explanation: Starting with

ax^2 + bx + c = 0

, factor out

a

a(x^2 + \frac{b}{a}x) + c = 0

. To complete the square, add and subtract

(\frac{b}{2a})^2 = \frac{b^2}{4a^2}

a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}) + c = 0

. This gives

a(x + \frac{b}{2a})^2 - a \cdot \frac{b^2}{4a^2} + c = 0

, so

a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c = 0

. Therefore

a(x + \frac{b}{2a})^2 = \frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a}

, giving

q = \frac{b^2 - 4ac}{4a}

. Choice B has the wrong sign. Choices C and D are missing the factor of

a

in the denominator.

Question 4

When completing the square for $x^2 - 8x + c = 0$ to obtain the form $(x - p)^2 = q$ , the value of $q$ is expressed in terms of $c$ . If the original equation has two real solutions, which condition must be satisfied?

$c < 16$ and $q = 16 - c$ (correct answer)
$c > 16$ and $q = c - 16$
$c < 4$ and $q = 4 - c$
$c > 4$ and $q = c - 4$

Explanation: To complete the square for

x^2 - 8x + c = 0

, we take half the coefficient of

x

and square it:

(-8/2)^2 = 16

. Rewriting:

x^2 - 8x + 16 - 16 + c = 0

, which gives

(x - 4)^2 - 16 + c = 0

, so

(x - 4)^2 = 16 - c

. Therefore

q = 16 - c

. For two real solutions, we need

q > 0

, so

16 - c > 0

, which means

c < 16

. Choice B incorrectly assumes

q = c - 16

and would require

c > 16

. Choices C and D use the wrong value when completing the square (4 instead of 16).

Question 5

After completing the square, the equation $2x^2 - 12x + 7 = 0$ can be written as $2(x - h)^2 = k$ . Using this form, what are the solutions to the original equation?

$x = 3 \pm \sqrt{\frac{11}{2}}$
$x = 3 \pm \frac{\sqrt{22}}{2}$ (correct answer)
$x = 6 \pm \sqrt{11}$
$x = 6 \pm \frac{\sqrt{11}}{2}$

Explanation: Starting with

2x^2 - 12x + 7 = 0

, factor out 2:

2(x^2 - 6x) + 7 = 0

. Complete the square:

2(x^2 - 6x + 9 - 9) + 7 = 0

, giving

2(x - 3)^2 - 18 + 7 = 0

, so

2(x - 3)^2 = 11

. Therefore

(x - 3)^2 = \frac{11}{2}

, which gives

x - 3 = \pm\sqrt{\frac{11}{2}}

, so

x = 3 \pm \sqrt{\frac{11}{2}}

. To rationalize:

\sqrt{\frac{11}{2}} = \frac{\sqrt{11}}{\sqrt{2}} = \frac{\sqrt{22}}{2}

, so

x = 3 \pm \frac{\sqrt{22}}{2}

. Choice A stops before rationalizing. Choice C uses wrong values (6 instead of 3). Choice D combines errors from both.

Question 6

Two students complete the square for $2x^2 + 8x - 3 = 0$ . Student A gets $2(x + 2)^2 = 11$ and Student B gets $(x + 2)^2 = \frac{11}{2}$ . Which statement about their work is correct?

Both forms are equivalent and correct, but Student A's form is more useful for graphing
Both forms are equivalent and correct, but Student B's form is more useful for solving (correct answer)
Only Student A is correct because the leading coefficient must remain factored out
Only Student B is correct because the coefficient of the squared term must equal 1

Explanation: Starting with

2x^2 + 8x - 3 = 0

: Factor out 2:

2(x^2 + 4x) - 3 = 0

. Complete the square:

2(x^2 + 4x + 4 - 4) - 3 = 0

, giving

2(x + 2)^2 - 8 - 3 = 0

, so

2(x + 2)^2 = 11

. This is Student A's form. Dividing both sides by 2 gives

(x + 2)^2 = \frac{11}{2}

, which is Student B's form. Both are mathematically equivalent and correct. However, Student B's form with coefficient 1 on the squared term makes it easier to solve by taking square roots directly:

x + 2 = \pm\sqrt{\frac{11}{2}}

, so

x = -2 \pm \sqrt{\frac{11}{2}}

. Choices C and D incorrectly suggest only one form is valid.

Question 7

Solve by completing the square: $2x^2+12x-10=0.$ (Be sure to divide by the leading coefficient first.)

$(x+3)^2=14$ , so $x=-3\pm \sqrt{14}$ . (correct answer)
$(x+6)^2=46$ , so $x=-6\pm \sqrt{46}$ .
$(x+3)^2=4$ , so $x=-3\pm 2$ (i.e., $x=-1$ or $x=-5$ ).
$(x-3)^2=14$ , so $x=3\pm \sqrt{14}$ .

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If

q

is positive, you get two real solutions; if

q = 0

, one solution; if

q

negative, two complex solutions

x = p \pm i\sqrt{|q|}

. The value of

q

immediately reveals the nature of solutions. To solve

2x^2 + 12x - 10 = 0

by completing the square: (1) Divide by 2:

x^2 + 6x - 5 = 0

. (2) Move constant:

x^2 + 6x = 5

. (3) Half of 6 is 3, square 9, add:

x^2 + 6x + 9 = 14

. (4)

(x + 3)^2 = 14

. (5)

x = -3 \pm \sqrt{14}

. The systematic process works every time! For deriving the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

q=4

instead of 14. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If

a \neq 1

, divide everything by

a

first to get

x^2

coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for

x

. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants: divide by

a

to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 8

Derive the quadratic formula by completing the square on the general quadratic equation $ax^2 + bx + c = 0 \quad (a \ne 0).$ Which expression for $x$ results?

$x=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}$
$x=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (correct answer)
$x=\dfrac{-b\pm\sqrt{b^2-2ac}}{a}$

Explanation: This question tests your ability to derive the quadratic formula by completing the square on the general form, showing it's not magic but systematic algebra. Completing the square transforms any quadratic

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If

q

is positive, you get two real solutions; if

q = 0

, one solution; if

q

negative, two complex solutions

x = p \pm i \sqrt{|q|}

. The value of

q

immediately reveals the nature of solutions. To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants: divide by

a

to get

x^2 + (b/a)x + (c/a) = 0

, then move

c/a

x^2 + (b/a)x = -c/a

, add

(b/(2a))^2

to both:

(x + b/(2a))^2 = -c/a + (b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square roots:

x + b/(2a) = \pm \sqrt{(b^2 - 4ac)/(4a^2)} = \pm \sqrt{b^2 - 4ac}/(2a)

, so

x = -b/(2a) \pm \sqrt{b^2 - 4ac}/(2a) = [-b \pm \sqrt{b^2 - 4ac}]/(2a)

. This derivation shows completing the square is the foundation! Choice C correctly derives the formula with the proper signs and discriminant. Choice A has a sign error in the discriminant, using

+4ac

instead of

-4ac

, which would mess up the nature of solutions. The systematic completing-the-square procedure: (1) If

a \ne 1

, divide everything by

a

first to get

x^2

coefficient

= 1

, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for

x

. To derive the quadratic formula, apply these steps symbolically as above—track the algebra carefully, and you'll see how the formula emerges every time! Amazing insight, well done!

Question 9

Solve by completing the square:

$x^2 - 8x + 5 = 0$

Write it in the form $(x - p)^2 = q$ and solve for $x$ .

$(x-4)^2=11$ , so $x=4 \pm \sqrt{11}$ (correct answer)
$(x+4)^2=11$ , so $x=-4 \pm \sqrt{11}$
$(x-4)^2=21$ , so $x=4 \pm \sqrt{21}$
$(x-8)^2=59$ , so $x=8 \pm \sqrt{59}$

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. To solve

x^2 - 8x + 5 = 0

by completing the square: (1) Move constant to right:

x^2 - 8x = -5

. (2) Take half of -8 to get -4, square it to get 16, add to both sides:

x^2 - 8x + 16 = -5 + 16 = 11

. (3) Factor left side:

(x - 4)^2 = 11

. (4) Take square roots:

x - 4 = \pm \sqrt{11}

. (5) Solve:

x = 4 \pm \sqrt{11}

. Choice A correctly completes the square to get

(x - 4)^2 = 11

and solves for both solutions. Choice B incorrectly factors as

(x + 4)^2

instead of

(x - 4)^2

—when the x coefficient is -8, half is -4, giving

(x - 4)^2

, not

(x + 4)^2

! The systematic procedure ensures you get the correct factored form by carefully tracking the sign of b/2.

Question 10

Solve using completing the square (do not factor): $x^2 + 12x + 20 = 0.$

$(x+6)^2=20$ , so $x=-6 \pm \sqrt{20}$
$(x-6)^2=16$ , so $x=6 \pm 4$ (i.e., $x=2$ or $x=10$ )
$(x+6)^2=16$ , so $x=-6 \pm 4$ (i.e., $x=-2$ or $x=-10$ ) (correct answer)
$(x+12)^2=124$ , so $x=-12 \pm \sqrt{124}$

Explanation: This question tests your ability to use completing the square to solve quadratic equations without factoring, even if it seems tempting. Completing the square transforms any quadratic

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If q is positive, you get two real solutions; if q = 0, one solution; if q negative, two complex solutions

x = p \pm i \sqrt{|q|}

. The value of q immediately reveals the nature of solutions. To solve

x^2 + 12x + 20 = 0

: (1) Move constant:

x^2 + 12x = -20

. (2) Half of 12 is 6, square 36, add both sides:

x^2 + 12x + 36 = -20 + 36 = 16

. (3)

(x + 6)^2 = 16

. (4)

x + 6 = \pm 4

. (5)

x = -6 \pm 4

, so x = -2 or -10. Perfect execution! Choice A correctly completes the square to get the proper form and solves for both solutions. Choice D flips the sign of p, likely by using half of -12 instead, which reverses the center. The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for x. Follow these 8 steps and you'll get the correct answer every time! You're mastering this!

Question 11

Derive the quadratic formula by completing the square on the general equation $ax^2 + bx + c = 0 \quad (a \neq 0).$ Which expression for $x$ results?

$x=\dfrac{b\pm \sqrt{b^2-4ac}}{2a}$
$x=\dfrac{-b\pm \sqrt{(b-2a)^2-4ac}}{a}$
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ (correct answer)
$x=\dfrac{-b\pm \sqrt{b^2+4ac}}{2a}$

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If

q

is positive, you get two real solutions; if

q = 0

, one solution; if

q

negative, two complex solutions

x = p \pm i\sqrt{|q|}

. The value of

q

immediately reveals the nature of solutions. To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants: divide by

a

to get

x^2 + (b/a)x + (c/a) = 0

, move constant:

x^2 + (b/a)x = -c/a

, add

(b/(2a))^2

to both:

(x + b/(2a))^2 = (b^2/(4a^2)) - (c/a) = (b^2 - 4ac)/(4a^2)

, take

\pm

square root:

x + b/(2a) = \pm \sqrt{(b^2 - 4ac)/(4a^2)} = \pm (\sqrt{b^2 - 4ac})/(2a)

, so

x = -b/(2a) \pm (\sqrt{b^2 - 4ac})/(2a) = [-b \pm \sqrt{b^2 - 4ac}] / (2a)

. The systematic process works every time! Choice C correctly derives the quadratic formula through completing the square on the general form. Choice A makes an error in the derivation of the quadratic formula during the completing square process with the general form, using

+4ac

instead of

-4ac

in the discriminant. When working with

a

b

c

as variables rather than numbers, track each step carefully—algebraic errors compound quickly in derivations! The systematic completing-the-square procedure: (1) If

a \neq 1

, divide everything by

a

first to get

x^2

coefficient

= 1

, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for

x

. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating

a

b

c

as constants: divide by

a

to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 12

Solve using completing the square: $x^2+2x+5=0.$ (Your completed-square form should look like $(x-p)^2=q$ .)

$(x+2)^2=-1$ , so $x=-2 \pm i$ .
$(x+1)^2=4$ , so $x=-1 \pm 2$ .
$(x+1)^2=-4$ , so $x=-1 \pm 2i$ . (correct answer)
$(x-1)^2=-4$ , so $x=1 \pm 2i$ .

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

x = p \pm i\sqrt{|q|}

. The value of q immediately reveals the nature of solutions. To solve

x^2 + 2x + 5 = 0

by completing the square: (1) Move constant:

x^2 + 2x = -5

. (2) Half of 2 is 1, square 1, add:

x^2 + 2x + 1 = -4

. (3)

(x + 1)^2 = -4

. (4)

x + 1 = \pm 2i

. (5)

x = -1 \pm 2i

. The systematic process works every time! For deriving the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants, following the same steps but keeping everything symbolic. Choice B correctly completes the square to get the proper form and solves to find both solutions accurately. Choice A incorrectly gets positive q=4, but it's negative for complex roots. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for x. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants: divide by a to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 13

Solve by completing the square and express the solutions in complex form (if needed):

$x^2-6x+13=0$

$(x-6)^2=-23$ , so $x=6\pm i\sqrt{23}$
$(x-3)^2=4$ , so $x=3\pm 2$
$(x+3)^2=-4$ , so $x=-3\pm 2i$
$(x-3)^2=-4$ , so $x=3\pm 2i$ (correct answer)

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots; if q is negative, you get complex solutions

x = p \pm i\sqrt{|q|}

. To solve

x^2 - 6x + 13 = 0

: (1) Move constant:

x^2 - 6x = -13

. (2) Take half of -6 to get -3, square it to get 9, add to both sides:

x^2 - 6x + 9 = -13 + 9 = -4

. (3) Factor:

(x - 3)^2 = -4

. (4) Take square roots:

x - 3 = \pm \sqrt{-4} = \pm 2i

. (5) Solve:

x = 3 \pm 2i

. Choice A correctly completes the square to get

(x - 3)^2 = -4

and recognizes that

\sqrt{-4} = 2i

, giving complex solutions. Choice B incorrectly factors as

(x + 3)^2

instead of

(x - 3)^2

—when the coefficient of x is -6, half is -3, so we get

(x - 3)^2

, not

(x + 3)^2

! The systematic procedure reveals when solutions are complex: whenever the right side after completing the square is negative.

Question 14

Solve by completing the square: $x^2-12x+20=0.$

$(x-6)^2=16$ , so $x=6\pm 4$ (i.e., $x=10$ or $x=2$ ). (correct answer)
$(x-12)^2=124$ , so $x=12\pm \sqrt{124}$ .
$(x+6)^2=16$ , so $x=-6\pm 4$ (i.e., $x=-2$ or $x=-10$ ).
$(x-6)^2=20$ , so $x=6\pm \sqrt{20}$ .

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

x = p \pm i\sqrt{|q|}

. The value of q immediately reveals the nature of solutions. To solve

x^2 - 12x + 20 = 0

by completing the square: (1) Move constant:

x^2 - 12x = -20

. (2) Half of -12 is -6, square 36, add:

x^2 - 12x + 36 = 16

. (3)

(x - 6)^2 = 16

. (4)

x - 6 = \pm 4

. (5)

x = 6 \pm 4

, so x = 10 or 2. The systematic process works every time! For deriving the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants, following the same steps but keeping everything symbolic. Choice A correctly completes the square to get the proper form and solves to find both solutions accurately. Choice C has the sign wrong for p, using +6 instead of -6 for negative b term. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for x. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants: divide by a to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 15

Solve by completing the square:

$x^2+12x+20=0$

Write it as $(x-p)^2=q$ and solve.

$(x+6)^2=16$ , so $x=-6\pm 4$ (correct answer)
$(x+6)^2=56$ , so $x=-6\pm\sqrt{56}$
$(x-6)^2=16$ , so $x=6\pm 4$
$(x+12)^2=124$ , so $x=-12\pm\sqrt{124}$

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. To solve

x^2 + 12x + 20 = 0

by completing the square: (1) Move constant to right:

x^2 + 12x = -20

. (2) Take half of 12 to get 6, square it to get 36, add to both sides:

x^2 + 12x + 36 = -20 + 36 = 16

. (3) Factor left side:

(x + 6)^2 = 16

. (4) Take square roots:

x + 6 = \pm 4

. (5) Solve:

x = -6 \pm 4

, so

x = -2

x = -10

. Choice A correctly completes the square to get

(x + 6)^2 = 16

and recognizes that

\sqrt{16} = 4

. Choice B incorrectly calculates

-20 + 36 = 56

instead of 16—always double-check your arithmetic! The systematic procedure gives you

x = -6 \pm 4

, which means

x = -2

x = -10

Question 16

Solve by completing the square: $x^2 + 4x - 5 = 0.$

$(x+4)^2=21$ , so $x=-4 \pm \sqrt{21}$
$(x+2)^2=1$ , so $x=-2 \pm 1$ (i.e., $x=-1$ or $x=-3$ )
$(x+2)^2=9$ , so $x=-2 \pm 3$ (i.e., $x=1$ or $x=-5$ ) (correct answer)
$(x-2)^2=9$ , so $x=2 \pm 3$ (i.e., $x=5$ or $x=-1$ )

Explanation: This question tests your ability to use completing the square to solve quadratic equations that factor nicely, but remember to use the method as instructed. Completing the square transforms any quadratic

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

x = p \pm i\sqrt{|q|}

. The value of q immediately reveals the nature of solutions. To solve

x^2 + 4x - 5 = 0

: (1) Move constant:

x^2 + 4x = 5

. (2) Half of 4 is 2, square 4, add both sides:

x^2 + 4x + 4 = 5 + 4 = 9

. (3)

(x + 2)^2 = 9

. (4)

x + 2 = \pm 3

. (5)

x = -2 \pm 3

, so

x = 1

-5

. You're nailing the steps! Choice A correctly completes the square to get the proper form and finds both integer solutions. Choice B miscalculates q as 1, perhaps by forgetting to add the square term to the right side properly. The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it:

(b/2)^2

, (4) Add to BOTH sides, (5) Factor left as

(x + b/2)^2

, (6) Simplify right, (7) Take

\pm

square root, (8) Solve for x. Follow these 8 steps and you'll get the correct answer every time! Keep up the great progress!

Question 17

Solve the equation by completing the square: $x^2+4x-5=0.$

$(x+4)^2=21$ , so $x=-4\pm \sqrt{21}$ .
$(x+2)^2=9$ , so $x=-2\pm 3$ (i.e., $x=1$ or $x=-5$ ). (correct answer)
$(x-2)^2=9$ , so $x=2\pm 3$ (i.e., $x=5$ or $x=-1$ ).
$(x+2)^2=1$ , so $x=-2\pm 1$ (i.e., $x=-1$ or $x=-3$ ).

Explanation: This question tests your ability to use completing the square to solve quadratic equations and, importantly, to derive the quadratic formula—showing it's not magic but comes from systematic algebra. Completing the square transforms any quadratic ax² + bx + c = 0 into the form (x - p)² = q, which is easy to solve by taking square roots: x - p = ±√q, giving x = p ± √q. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If q is positive, you get two real solutions; if q = 0, one solution; if q negative, two complex solutions x = p ± i√|q|. The value of q immediately reveals the nature of solutions. To solve x² + 4x - 5 = 0 by completing the square: (1) Move constant: x² + 4x = 5. (2) Half of 4 is 2, square 4, add: x² + 4x + 4 = 9. (3) (x + 2)² = 9. (4) x + 2 = ±3. (5) x = -2 ± 3, so x = 1 or -5. The systematic process works every time! For deriving the quadratic formula, complete the square on ax² + bx + c = 0 treating a, b, c as constants, following the same steps but keeping everything symbolic. Choice B correctly completes the square to get the proper form and solves to find both solutions accurately. Choice A gets q=1 instead of 9, likely forgetting to add 4 to the right side correctly. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x² coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)², (4) Add to BOTH sides, (5) Factor left as (x + b/2)², (6) Simplify right, (7) Take ± square root, (8) Solve for x. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on ax² + bx + c = 0 treating a, b, c as constants: divide by a to get x² + (b/a)x + (c/a) = 0, then complete square on x² + (b/a)x by adding (b/(2a))², which gives (x + b/(2a))² = (b² - 4ac)/(4a²), then take square root and solve to get x = (-b ± √(b² - 4ac)) / (2a). This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 18

Use completing the square to solve:

$3x^2+18x+15=0$

(First divide by the leading coefficient, then complete the square.)

$(x+3)^2=4$ , so $x=-3\pm 2$ (correct answer)
$(x+3)^2=-4$ , so $x=-3\pm 2i$
$(x-3)^2=4$ , so $x=3\pm 2$
$(x+6)^2=13$ , so $x=-6\pm\sqrt{13}$

a x^2 + b x + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

. To solve

3x^2 + 18x + 15 = 0

by completing the square: (1) Divide by 3 to get

x^2 + 6x + 5 = 0

. (2) Move constant:

x^2 + 6x = -5

. (3) Take half of 6 to get 3, square it to get 9, add to both sides:

x^2 + 6x + 9 = -5 + 9 = 4

. (4) Factor:

(x + 3)^2 = 4

. (5) Take square roots:

x + 3 = \pm 2

. (6) Solve:

x = -3 \pm 2

, so

x = -1

x = -5

. Choice A correctly completes the square after dividing by the leading coefficient to get

(x + 3)^2 = 4

. Choice B incorrectly gets a negative value on the right side—when you add 9 to -5, you get positive 4, not negative 4! The systematic procedure requires first dividing by the leading coefficient when it's not 1.

Question 19

Solve the quadratic equation by completing the square: $x^2+8x+7=0$ . Which choice gives the correct completed-square form and solutions?

$(x+4)^2=9$ , so $x=-4\pm 3$ (i.e., $x=-1$ or $x=-7$ ) (correct answer)
$(x+8)^2=57$ , so $x=-8\pm \sqrt{57}$
$(x+4)^2=7$ , so $x=-4\pm \sqrt{7}$
$(x-4)^2=9$ , so $x=4\pm 3$ (i.e., $x=1$ or $x=7$ )

Explanation: This question tests your ability to use completing the square to solve quadratic equations and understand how it leads to solutions, much like deriving the quadratic formula shows it's rooted in algebra. Completing the square transforms ax² + bx + c = 0 into (x - p)² = q, solved by x = p ± √q, working for all quadratics including those with real or complex roots; if q > 0, two real solutions; q = 0, one; q < 0, two complex. For x² + 8x + 7 = 0, move 7: x² + 8x = -7, add (8/2)² = 16: (x + 4)² = 9, then x + 4 = ±3, so x = -1 or -7—great job following the steps! Choice A correctly completes the square to (x + 4)² = 9 and finds both solutions accurately. Choice B mistakenly uses half of 8 as 8 instead of 4, leading to an incorrect perfect square term and wrong q value—remember, it's always half the coefficient squared! The systematic process is: move constant, add (b/2)² to both sides, factor, take ± square root, solve—practice this and you'll master it every time! Deriving the quadratic formula follows similarly on ax² + bx + c = 0, yielding x = [-b ± √(b² - 4ac)] / (2a), proving the formula's algebraic foundation.

Question 20

Solve using the method of completing the square: $x^2 + 6x - 2 = 0.$ First rewrite it in the form $(x - p)^2 = q$ .

$(x+3)^2=7$ , so $x=-3 \pm \sqrt{7}$ .
$(x-3)^2=11$ , so $x=3 \pm \sqrt{11}$ .
$(x+6)^2=38$ , so $x=-6 \pm \sqrt{38}$ .
$(x+3)^2=11$ , so $x=-3 \pm \sqrt{11}$ . (correct answer)

ax^2 + bx + c = 0

into the form

(x - p)^2 = q

, which is easy to solve by taking square roots:

x - p = \pm \sqrt{q}

, giving

x = p \pm \sqrt{q}

x = p \pm i\sqrt{|q|}

. The value of q immediately reveals the nature of solutions. To solve

x^2 + 6x - 2 = 0

by completing the square: (1) Move constant to right:

x^2 + 6x = 2

. (2) Take half of 6 to get 3, square it to get 9, add to both sides:

x^2 + 6x + 9 = 2 + 9 = 11

. (3) Factor left side as perfect square:

(x + 3)^2 = 11

. (4) Take square roots:

x + 3 = \pm \sqrt{11}

. (5) Solve:

x = -3 \pm \sqrt{11}

. The systematic process works every time! For deriving the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants, following the same steps but keeping everything symbolic. Choice B correctly completes the square to get the proper form and solves to find both solutions accurately. Choice A calculates (b/2)^2 incorrectly: half of 6 is 3, and 3 squared is 9, leading to +9 on right making 11, not 7. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x^2 coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x + b/2)^2, (6) Simplify right, (7) Take ± square root, (8) Solve for x. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on

ax^2 + bx + c = 0

treating a, b, c as constants: divide by a to get

x^2 + (b/a)x + (c/a) = 0

, then complete square on

x^2 + (b/a)x

by adding

(b/(2a))^2

, which gives

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

, then take square root and solve to get

x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)

. This derivation shows completing the square is the fundamental method underlying the quadratic formula!