Algebra 2 Quiz: Complete The Square To Find Solutions
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Complete The Square To Find SolutionsQuestion 1 of 20

Transform and solve by completing the square:

x2+10x3=0x^2+10x-3=0

Write it as (xp)2=q(x-p)^2=q and then solve for xx.

(x+5)2=22(x+5)^2=22, so x=5±22x=-5\pm\sqrt{22}
(x+5)2=28(x+5)^2=28, so x=5±27x=-5\pm 2\sqrt{7}
(x5)2=28(x-5)^2=28, so x=5±27x=5\pm 2\sqrt{7}
(x+10)2=103(x+10)^2=103, so x=10±103x=-10\pm\sqrt{103}
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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

Transform and solve by completing the square:

x2+10x3=0x^2+10x-3=0

Write it as (xp)2=q(x-p)^2=q and then solve for xx.

  1. (x+5)2=22(x+5)^2=22, so x=5±22x=-5\pm\sqrt{22}
  2. (x+5)2=28(x+5)^2=28, so x=5±27x=-5\pm 2\sqrt{7} (correct answer)
  3. (x5)2=28(x-5)^2=28, so x=5±27x=5\pm 2\sqrt{7}
  4. (x+10)2=103(x+10)^2=103, so x=10±103x=-10\pm\sqrt{103}
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. To solve x² + 10x - 3 = 0 by completing the square: (1) Move constant to right: x² + 10x = 3. (2) Take half of 10 to get 5, square it to get 25, add to both sides: x² + 10x + 25 = 3 + 25 = 28. (3) Factor left side: (x + 5)² = 28. (4) Take square roots: x + 5 = ±√28 = ±2√7. (5) Solve: x = -5 ± 2√7. Choice A correctly completes the square to get (x + 5)² = 28 and identifies that √28 = √(4·7) = 2√7. Choice C makes an arithmetic error: adding 25 to 3 gives 28, not 22—always double-check your arithmetic when completing the square! The systematic procedure works perfectly when you carefully add (b/2)² = 25 to both sides.

Question 2

Solve by completing the square:

x2+12x+20=0x^2+12x+20=0

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

  1. (x+6)2=16(x+6)^2=16, so x=6±4x=-6\pm 4 (correct answer)
  2. (x+6)2=56(x+6)^2=56, so x=6±56x=-6\pm\sqrt{56}
  3. (x6)2=16(x-6)^2=16, so x=6±4x=6\pm 4
  4. (x+12)2=124(x+12)^2=124, so x=12±124x=-12\pm\sqrt{124}
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. To solve x2+12x+20=0x^2 + 12x + 20 = 0 by completing the square: (1) Move constant to right: x2+12x=20x^2 + 12x = -20. (2) Take half of 12 to get 6, square it to get 36, add to both sides: x2+12x+36=20+36=16x^2 + 12x + 36 = -20 + 36 = 16. (3) Factor left side: (x+6)2=16(x + 6)^2 = 16. (4) Take square roots: x+6=±4x + 6 = \pm 4. (5) Solve: x=6±4x = -6 \pm 4, so x=2x = -2 or x=10x = -10. Choice A correctly completes the square to get (x+6)2=16(x + 6)^2 = 16 and recognizes that 16=4\sqrt{16} = 4. Choice B incorrectly calculates 20+36=56-20 + 36 = 56 instead of 16—always double-check your arithmetic! The systematic procedure gives you x=6±4x = -6 \pm 4, which means x=2x = -2 or x=10x = -10.

Question 3

Solve the quadratic equation by completing the square: x2+8x+7=0.x^2 + 8x + 7 = 0. Write your work in the form (xp)2=q(x - p)^2 = q before solving.

  1. (x+4)2=9(x+4)^2=9, so x=4±3x=-4 \pm 3 (i.e., x=1x=-1 or x=7x=-7). (correct answer)
  2. (x+8)2=57(x+8)^2=57, so x=8±57x=-8 \pm \sqrt{57}.
  3. (x+4)2=7(x+4)^2=7, so x=4±7x=-4 \pm \sqrt{7}.
  4. (x4)2=9(x-4)^2=9, so x=4±3x=4 \pm 3 (i.e., x=1x=1 or x=7x=7).
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If qq is positive, you get two real solutions; if q=0q = 0, one solution; if qq negative, two complex solutions x=p±iqx = p \pm i \sqrt{|q|}. The value of qq immediately reveals the nature of solutions. To solve x2+8x+7=0x^2 + 8x + 7 = 0 by completing the square: (1) Move constant to right: x2+8x=7x^2 + 8x = -7. (2) Take half of 8 to get 4, square it to get 16, add to both sides: x2+8x+16=7+16=9x^2 + 8x + 16 = -7 + 16 = 9. (3) Factor left side as perfect square: (x+4)2=9(x + 4)^2 = 9. (4) Take square roots: x+4=±3x + 4 = \pm 3. (5) Solve: x=4±3x = -4 \pm 3, so x=1x = -1 or x=7x = -7. The systematic process works every time! For deriving the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc 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(b/2)^2 incorrectly: half of 8 is 4, and 4 squared is 16, not leading to (x+8)2=57(x+8)^2=57. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a1a \neq 1, divide everything by aa first to get x2x^2 coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)2(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(x + b/2)^2, (6) Simplify right, (7) Take ±\pm square root, (8) Solve for xx. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc as constants: divide by aa to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then complete square on x2+(b/a)xx^2 + (b/a)x by adding (b/(2a))2(b/(2a))^2, which gives (x+b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square root and solve to get x=(b±b24ac)/(2a)x = (-b \pm \sqrt{b^2 - 4ac}) / (2a). This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 4

Solve by completing the square:

x28x+5=0x^2 - 8x + 5 = 0

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

  1. (x4)2=11(x-4)^2=11, so x=4±11x=4 \pm \sqrt{11} (correct answer)
  2. (x+4)2=11(x+4)^2=11, so x=4±11x=-4 \pm \sqrt{11}
  3. (x4)2=21(x-4)^2=21, so x=4±21x=4 \pm \sqrt{21}
  4. (x8)2=59(x-8)^2=59, so x=8±59x=8 \pm \sqrt{59}
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. To solve x28x+5=0x^2 - 8x + 5 = 0 by completing the square: (1) Move constant to right: x28x=5x^2 - 8x = -5. (2) Take half of -8 to get -4, square it to get 16, add to both sides: x28x+16=5+16=11x^2 - 8x + 16 = -5 + 16 = 11. (3) Factor left side: (x4)2=11(x - 4)^2 = 11. (4) Take square roots: x4=±11x - 4 = \pm \sqrt{11}. (5) Solve: x=4±11x = 4 \pm \sqrt{11}. Choice A correctly completes the square to get (x4)2=11(x - 4)^2 = 11 and solves for both solutions. Choice B incorrectly factors as (x+4)2(x + 4)^2 instead of (x4)2(x - 4)^2—when the x coefficient is -8, half is -4, giving (x4)2(x - 4)^2, not (x+4)2(x + 4)^2! The systematic procedure ensures you get the correct factored form by carefully tracking the sign of b/2.

Question 5

Derive the quadratic formula by completing the square on the general quadratic equation ax2+bx+c=0(a0).ax^2 + bx + c = 0 \quad (a \ne 0). Which expression for xx results?

  1. x=b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}
  2. x=b±b24ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}
  3. x=b±b24ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} (correct answer)
  4. x=b±b22acax=\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 ax2+bx+c=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If qq is positive, you get two real solutions; if q=0q = 0, one solution; if qq negative, two complex solutions x=p±iqx = p \pm i \sqrt{|q|}. The value of qq immediately reveals the nature of solutions. To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc as constants: divide by aa to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then move c/ac/a: x2+(b/a)x=c/ax^2 + (b/a)x = -c/a, add (b/(2a))2(b/(2a))^2 to both: (x+b/(2a))2=c/a+(b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = -c/a + (b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square roots: x+b/(2a)=±(b24ac)/(4a2)=±b24ac/(2a)x + b/(2a) = \pm \sqrt{(b^2 - 4ac)/(4a^2)} = \pm \sqrt{b^2 - 4ac}/(2a), so x=b/(2a)±b24ac/(2a)=[b±b24ac]/(2a)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+4ac instead of 4ac-4ac, which would mess up the nature of solutions. The systematic completing-the-square procedure: (1) If a1a \ne 1, divide everything by aa first to get x2x^2 coefficient =1= 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)2(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(x + b/2)^2, (6) Simplify right, (7) Take ±\pm square root, (8) Solve for xx. 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 6

Derive the quadratic formula by completing the square on the general equation ax2+bx+c=0(a0).ax^2 + bx + c = 0 \quad (a \neq 0). Which expression for xx results?

  1. x=b±b24ac2ax=\dfrac{b\pm \sqrt{b^2-4ac}}{2a}
  2. x=b±(b2a)24acax=\dfrac{-b\pm \sqrt{(b-2a)^2-4ac}}{a}
  3. x=b±b24ac2ax=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} (correct answer)
  4. x=b±b2+4ac2ax=\dfrac{-b\pm \sqrt{b^2+4ac}}{2a}
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If qq is positive, you get two real solutions; if q=0q = 0, one solution; if qq negative, two complex solutions x=p±iqx = p \pm i\sqrt{|q|}. The value of qq immediately reveals the nature of solutions. To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc as constants: divide by aa to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, move constant: x2+(b/a)x=c/ax^2 + (b/a)x = -c/a, add (b/(2a))2(b/(2a))^2 to both: (x+b/(2a))2=(b2/(4a2))(c/a)=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2/(4a^2)) - (c/a) = (b^2 - 4ac)/(4a^2), take ±\pm square root: x+b/(2a)=±(b24ac)/(4a2)=±(b24ac)/(2a)x + b/(2a) = \pm \sqrt{(b^2 - 4ac)/(4a^2)} = \pm (\sqrt{b^2 - 4ac})/(2a), so x=b/(2a)±(b24ac)/(2a)=[b±b24ac]/(2a)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+4ac instead of 4ac-4ac in the discriminant. When working with aa, bb, cc as variables rather than numbers, track each step carefully—algebraic errors compound quickly in derivations! The systematic completing-the-square procedure: (1) If a1a \neq 1, divide everything by aa first to get x2x^2 coefficient =1= 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)2(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(x + b/2)^2, (6) Simplify right, (7) Take ±\pm square root, (8) Solve for xx. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc as constants: divide by aa to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then complete square on x2+(b/a)xx^2 + (b/a)x by adding (b/(2a))2(b/(2a))^2, which gives (x+b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square root and solve to get x=(b±b24ac)/(2a)x = (-b \pm \sqrt{b^2 - 4ac}) / (2a). This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 7

Solve using completing the square: x2+2x+5=0.x^2+2x+5=0. (Your completed-square form should look like (xp)2=q(x-p)^2=q.)

  1. (x+2)2=1(x+2)^2=-1, so x=2±ix=-2 \pm i.
  2. (x+1)2=4(x+1)^2=4, so x=1±2x=-1 \pm 2.
  3. (x+1)2=4(x+1)^2=-4, so x=1±2ix=-1 \pm 2i. (correct answer)
  4. (x1)2=4(x-1)^2=-4, so x=1±2ix=1 \pm 2i.
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = 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±iqx = p \pm i\sqrt{|q|}. The value of q immediately reveals the nature of solutions. To solve x2+2x+5=0x^2 + 2x + 5 = 0 by completing the square: (1) Move constant: x2+2x=5x^2 + 2x = -5. (2) Half of 2 is 1, square 1, add: x2+2x+1=4x^2 + 2x + 1 = -4. (3) (x+1)2=4(x + 1)^2 = -4. (4) x+1=±2ix + 1 = \pm 2i. (5) x=1±2ix = -1 \pm 2i. The systematic process works every time! For deriving the quadratic formula, complete the square on ax2+bx+c=0ax^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(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(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 ax2+bx+c=0ax^2 + bx + c = 0 treating a, b, c as constants: divide by a to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then complete square on x2+(b/a)xx^2 + (b/a)x by adding (b/(2a))2(b/(2a))^2, which gives (x+b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square root and solve to get x=(b±b24ac)/(2a)x = (-b \pm \sqrt{b^2 - 4ac}) / (2a). This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 8

Solve the equation by completing the square:

x2+4x+13=0x^2 + 4x + 13 = 0

Express solutions in the form a±bia \pm bi.

  1. (x+2)2=9(x + 2)^2 = -9, so x=2±3ix = -2 \pm 3i (correct answer)
  2. (x2)2=9(x - 2)^2 = -9, so x=2±3ix = 2 \pm 3i
  3. (x+2)2=9(x + 2)^2 = 9, so x=2±3x = -2 \pm 3
  4. (x+4)2=3(x + 4)^2 = -3, so x=4±i3x = -4 \pm i \sqrt{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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots; if q is negative, you get complex solutions x=p±iqx = p \pm i \sqrt{|q|}. To solve x2+4x+13=0x^2 + 4x + 13 = 0: (1) Move constant: x2+4x=13x^2 + 4x = -13. (2) Take half of 4 to get 2, square it to get 4, add to both sides: x2+4x+4=13+4=9x^2 + 4x + 4 = -13 + 4 = -9. (3) Factor: (x+2)2=9(x + 2)^2 = -9. (4) Take square roots: x+2=±9=±3ix + 2 = \pm \sqrt{-9} = \pm 3i. (5) Solve: x=2±3ix = -2 \pm 3i. Choice A correctly completes the square to get (x+2)2=9(x + 2)^2 = -9 and recognizes that 9=3i\sqrt{-9} = 3i, giving complex solutions. Choice C incorrectly calculates the right side as positive 9 instead of negative 9—remember 13+4=9-13 + 4 = -9, not +9+9! The negative value on the right side after completing the square signals that you'll get complex solutions.

Question 9

Complete the square on the general quadratic equation to derive the quadratic formula. Starting from

ax2+bx+c=0(a0),ax^2+bx+c=0 \quad (a\ne 0),

which expression for xx results from completing the square?

  1. x=b±b24ac2ax=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}
  2. x=b±b2+4ac2ax=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}
  3. x=b±b22acax=\dfrac{-b\pm\sqrt{b^2-2ac}}{a}
  4. x=b±b24ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} (correct answer)
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating a, b, c as constants: divide by a to get x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0, then complete square on x2+baxx^2 + \frac{b}{a}x by adding (b2a)2\left(\frac{b}{2a}\right)^2, which gives (x+b2a)2=b24ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}, then take square root and solve to get x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Choice C correctly shows the quadratic formula with the proper negative sign on b and the discriminant b24acb^2 - 4ac. Choice A incorrectly has b2+4acb^2 + 4ac under the square root—the discriminant must be b24acb^2 - 4ac, not b2+4acb^2 + 4ac! The systematic derivation by completing the square reveals why the quadratic formula has exactly this form, including the crucial minus sign in the discriminant.

Question 10

Complete the square to solve: x2+6x2=0.x^2 + 6x - 2 = 0. Write the equation in the form (xp)2=q(x - p)^2 = q and then give the solutions.

  1. (x+6)2=38(x + 6)^2 = 38, so x=6±38x = -6 \pm \sqrt{38}
  2. (x3)2=11(x - 3)^2 = 11, so x=3±11x = 3 \pm \sqrt{11}
  3. (x+3)2=7(x + 3)^2 = 7, so x=3±7x = -3 \pm \sqrt{7}
  4. (x+3)2=11(x + 3)^2 = 11, so x=3±11x = -3 \pm \sqrt{11} (correct answer)
Explanation: This question tests your ability to use completing the square to solve quadratic equations, highlighting the importance of the sign when moving terms. Completing the square transforms any quadratic ax2+bx+c=0a x^2 + b x + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = 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±iqx = p \pm i \sqrt{|q|}. The value of q immediately reveals the nature of solutions. To solve x2+6x2=0x^2 + 6x - 2 = 0: (1) Move constant: x2+6x=2x^2 + 6x = 2. (2) Half of 6 is 3, square 9, add both sides: x2+6x+9=2+9=11x^2 + 6x + 9 = 2 + 9 = 11. (3) (x+3)2=11(x + 3)^2 = 11. (4) x=3±11x = -3 \pm \sqrt{11}. The process is straightforward when you move terms correctly! Choice A correctly completes the square to get the proper form and solves to find both solutions accurately. Choice C uses the wrong sign for p, perhaps by taking half of -6 instead of 6, leading to an incorrect center. The systematic completing-the-square procedure: (1) If a ≠ 1, divide everything by a first to get x2x^2 coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)2(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(x + b/2)^2, (6) Simplify right, (7) Take ± square root, (8) Solve for x. Follow these 8 steps and you'll get the correct answer every time! Awesome work!

Question 11

Solve using the method of completing the square: x2+6x2=0.x^2 + 6x - 2 = 0. First rewrite it in the form (xp)2=q(x - p)^2 = q.

  1. (x+3)2=7(x+3)^2=7, so x=3±7x=-3 \pm \sqrt{7}.
  2. (x3)2=11(x-3)^2=11, so x=3±11x=3 \pm \sqrt{11}.
  3. (x+6)2=38(x+6)^2=38, so x=6±38x=-6 \pm \sqrt{38}.
  4. (x+3)2=11(x+3)^2=11, so x=3±11x=-3 \pm \sqrt{11}. (correct answer)
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = 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±iqx = p \pm i\sqrt{|q|}. The value of q immediately reveals the nature of solutions. To solve x2+6x2=0x^2 + 6x - 2 = 0 by completing the square: (1) Move constant to right: x2+6x=2x^2 + 6x = 2. (2) Take half of 6 to get 3, square it to get 9, add to both sides: x2+6x+9=2+9=11x^2 + 6x + 9 = 2 + 9 = 11. (3) Factor left side as perfect square: (x+3)2=11(x + 3)^2 = 11. (4) Take square roots: x+3=±11x + 3 = \pm \sqrt{11}. (5) Solve: x=3±11x = -3 \pm \sqrt{11}. The systematic process works every time! For deriving the quadratic formula, complete the square on ax2+bx+c=0ax^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 ax2+bx+c=0ax^2 + bx + c = 0 treating a, b, c as constants: divide by a to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then complete square on x2+(b/a)xx^2 + (b/a)x by adding (b/(2a))2(b/(2a))^2, which gives (x+b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square root and solve to get x=(b±b24ac)/(2a)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 the equation by completing the square: x2+4x5=0.x^2+4x-5=0.

  1. (x+4)2=21(x+4)^2=21, so x=4±21x=-4\pm \sqrt{21}.
  2. (x+2)2=9(x+2)^2=9, so x=2±3x=-2\pm 3 (i.e., x=1x=1 or x=5x=-5). (correct answer)
  3. (x2)2=9(x-2)^2=9, so x=2±3x=2\pm 3 (i.e., x=5x=5 or x=1x=-1).
  4. (x+2)2=1(x+2)^2=1, so x=2±1x=-2\pm 1 (i.e., x=1x=-1 or x=3x=-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 13

Use completing the square to solve:

3x2+18x+15=03x^2+18x+15=0

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

  1. (x+3)2=4(x+3)^2=4, so x=3±2x=-3\pm 2 (correct answer)
  2. (x+3)2=4(x+3)^2=-4, so x=3±2ix=-3\pm 2i
  3. (x3)2=4(x-3)^2=4, so x=3±2x=3\pm 2
  4. (x+6)2=13(x+6)^2=13, so x=6±13x=-6\pm\sqrt{13}
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=0a x^2 + b x + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. To solve 3x2+18x+15=03x^2 + 18x + 15 = 0 by completing the square: (1) Divide by 3 to get x2+6x+5=0x^2 + 6x + 5 = 0. (2) Move constant: x2+6x=5x^2 + 6x = -5. (3) Take half of 6 to get 3, square it to get 9, add to both sides: x2+6x+9=5+9=4x^2 + 6x + 9 = -5 + 9 = 4. (4) Factor: (x+3)2=4(x + 3)^2 = 4. (5) Take square roots: x+3=±2x + 3 = \pm 2. (6) Solve: x=3±2x = -3 \pm 2, so x=1x = -1 or x=5x = -5. Choice A correctly completes the square after dividing by the leading coefficient to get (x+3)2=4(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 14

When deriving the quadratic formula from ax2+bx+c=0ax^2 + bx + c = 0 using completing the square, the intermediate step a(x+b2a)2=b24ac4aa(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a} is reached. What is the next step that leads directly to the quadratic formula?

  1. Expand the left side, then rearrange to solve for xx
  2. Take the square root of both sides, then subtract b2a\frac{b}{2a} from both sides
  3. Divide both sides by aa, then take the square root of both sides (correct answer)
  4. Factor the right side, then divide both sides by 4a4a
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+b2a)2=b24ac4aa(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a}, the most direct path is to divide both sides by aa first, giving you (x+b2a)2=b24ac4a2(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}. Then take the square root of both sides: x+b2a=±b24ac2ax + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}. Finally, subtract b2a\frac{b}{2a} to get x=b±b24ac2ax = \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 aa, you get x+b2a=±b24ac2ax + \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 b24acb^2 - 4ac, but this expression (the discriminant) generally can't be factored in a useful way, and dividing by 4a4a 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 15

Consider the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 where a0a \neq 0. When this equation is transformed by completing the square to the form a(xp)2=qa(x - p)^2 = q, which expression correctly represents qq in terms of aa, bb, and cc?

  1. q=b24ac4aq = \frac{b^2 - 4ac}{4a} (correct answer)
  2. q=4acb24aq = \frac{4ac - b^2}{4a}
  3. q=b24ac4q = \frac{b^2 - 4ac}{4}
  4. q=4acb24q = \frac{4ac - b^2}{4}
Explanation: Starting with ax2+bx+c=0ax^2 + bx + c = 0, factor out aa: a(x2+bax)+c=0a(x^2 + \frac{b}{a}x) + c = 0. To complete the square, add and subtract (b2a)2=b24a2(\frac{b}{2a})^2 = \frac{b^2}{4a^2}: a(x2+bax+b24a2b24a2)+c=0a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}) + c = 0. This gives a(x+b2a)2ab24a2+c=0a(x + \frac{b}{2a})^2 - a \cdot \frac{b^2}{4a^2} + c = 0, so a(x+b2a)2b24a+c=0a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c = 0. Therefore a(x+b2a)2=b24ac=b24ac4aa(x + \frac{b}{2a})^2 = \frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a}, giving q=b24ac4aq = \frac{b^2 - 4ac}{4a}. Choice B has the wrong sign. Choices C and D are missing the factor of aa in the denominator.

Question 16

When completing the square for x28x+c=0x^2 - 8x + c = 0 to obtain the form (xp)2=q(x - p)^2 = q, the value of qq is expressed in terms of cc. If the original equation has two real solutions, which condition must be satisfied?

  1. c<16c < 16 and q=16cq = 16 - c (correct answer)
  2. c>16c > 16 and q=c16q = c - 16
  3. c<4c < 4 and q=4cq = 4 - c
  4. c>4c > 4 and q=c4q = c - 4
Explanation: To complete the square for x28x+c=0x^2 - 8x + c = 0, we take half the coefficient of xx and square it: (8/2)2=16(-8/2)^2 = 16. Rewriting: x28x+1616+c=0x^2 - 8x + 16 - 16 + c = 0, which gives (x4)216+c=0(x - 4)^2 - 16 + c = 0, so (x4)2=16c(x - 4)^2 = 16 - c. Therefore q=16cq = 16 - c. For two real solutions, we need q>0q > 0, so 16c>016 - c > 0, which means c<16c < 16. Choice B incorrectly assumes q=c16q = c - 16 and would require c>16c > 16. Choices C and D use the wrong value when completing the square (4 instead of 16).

Question 17

After completing the square, the equation 2x212x+7=02x^2 - 12x + 7 = 0 can be written as 2(xh)2=k2(x - h)^2 = k. Using this form, what are the solutions to the original equation?

  1. x=3±112x = 3 \pm \sqrt{\frac{11}{2}}
  2. x=3±222x = 3 \pm \frac{\sqrt{22}}{2} (correct answer)
  3. x=6±11x = 6 \pm \sqrt{11}
  4. x=6±112x = 6 \pm \frac{\sqrt{11}}{2}
Explanation: Starting with 2x212x+7=02x^2 - 12x + 7 = 0, factor out 2: 2(x26x)+7=02(x^2 - 6x) + 7 = 0. Complete the square: 2(x26x+99)+7=02(x^2 - 6x + 9 - 9) + 7 = 0, giving 2(x3)218+7=02(x - 3)^2 - 18 + 7 = 0, so 2(x3)2=112(x - 3)^2 = 11. Therefore (x3)2=112(x - 3)^2 = \frac{11}{2}, which gives x3=±112x - 3 = \pm\sqrt{\frac{11}{2}}, so x=3±112x = 3 \pm \sqrt{\frac{11}{2}}. To rationalize: 112=112=222\sqrt{\frac{11}{2}} = \frac{\sqrt{11}}{\sqrt{2}} = \frac{\sqrt{22}}{2}, so x=3±222x = 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 18

Two students complete the square for 2x2+8x3=02x^2 + 8x - 3 = 0. Student A gets 2(x+2)2=112(x + 2)^2 = 11 and Student B gets (x+2)2=112(x + 2)^2 = \frac{11}{2}. Which statement about their work is correct?

  1. Both forms are equivalent and correct, but Student A's form is more useful for graphing
  2. Both forms are equivalent and correct, but Student B's form is more useful for solving (correct answer)
  3. Only Student A is correct because the leading coefficient must remain factored out
  4. Only Student B is correct because the coefficient of the squared term must equal 1
Explanation: Starting with 2x2+8x3=02x^2 + 8x - 3 = 0: Factor out 2: 2(x2+4x)3=02(x^2 + 4x) - 3 = 0. Complete the square: 2(x2+4x+44)3=02(x^2 + 4x + 4 - 4) - 3 = 0, giving 2(x+2)283=02(x + 2)^2 - 8 - 3 = 0, so 2(x+2)2=112(x + 2)^2 = 11. This is Student A's form. Dividing both sides by 2 gives (x+2)2=112(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=±112x + 2 = \pm\sqrt{\frac{11}{2}}, so x=2±112x = -2 \pm \sqrt{\frac{11}{2}}. Choices C and D incorrectly suggest only one form is valid.

Question 19

Solve by completing the square: 2x2+12x10=0.2x^2+12x-10=0. (Be sure to divide by the leading coefficient first.)

  1. (x+3)2=14(x+3)^2=14, so x=3±14x=-3\pm \sqrt{14}. (correct answer)
  2. (x+6)2=46(x+6)^2=46, so x=6±46x=-6\pm \sqrt{46}.
  3. (x+3)2=4(x+3)^2=4, so x=3±2x=-3\pm 2 (i.e., x=1x=-1 or x=5x=-5).
  4. (x3)2=14(x-3)^2=14, so x=3±14x=3\pm \sqrt{14}.
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=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = p \pm \sqrt{q}. This method works for ALL quadratics—even those that don't factor and even those with complex solutions! If qq is positive, you get two real solutions; if q=0q = 0, one solution; if qq negative, two complex solutions x=p±iqx = p \pm i\sqrt{|q|}. The value of qq immediately reveals the nature of solutions. To solve 2x2+12x10=02x^2 + 12x - 10 = 0 by completing the square: (1) Divide by 2: x2+6x5=0x^2 + 6x - 5 = 0. (2) Move constant: x2+6x=5x^2 + 6x = 5. (3) Half of 6 is 3, square 9, add: x2+6x+9=14x^2 + 6x + 9 = 14. (4) (x+3)2=14(x + 3)^2 = 14. (5) x=3±14x = -3 \pm \sqrt{14}. The systematic process works every time! For deriving the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc 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 forgets to add the correct amount after dividing, leading to wrong q=4q=4 instead of 14. This perfect square term is crucial—getting it wrong throws off everything! The systematic completing-the-square procedure: (1) If a1a \neq 1, divide everything by aa first to get x2x^2 coefficient = 1, (2) Move constant to right, (3) Take half the x-coefficient, square it: (b/2)2(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(x + b/2)^2, (6) Simplify right, (7) Take ±\pm square root, (8) Solve for xx. Follow these 8 steps and you'll get correct answer every time! To derive the quadratic formula, complete the square on ax2+bx+c=0ax^2 + bx + c = 0 treating aa, bb, cc as constants: divide by aa to get x2+(b/a)x+(c/a)=0x^2 + (b/a)x + (c/a) = 0, then complete square on x2+(b/a)xx^2 + (b/a)x by adding (b/(2a))2(b/(2a))^2, which gives (x+b/(2a))2=(b24ac)/(4a2)(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), then take square root and solve to get x=(b±b24ac)/(2a)x = (-b \pm \sqrt{b^2 - 4ac}) / (2a). This derivation shows completing the square is the fundamental method underlying the quadratic formula!

Question 20

Solve using completing the square (do not factor): x2+12x+20=0.x^2 + 12x + 20 = 0.

  1. (x+6)2=20(x+6)^2=20, so x=6±20x=-6 \pm \sqrt{20}
  2. (x6)2=16(x-6)^2=16, so x=6±4x=6 \pm 4 (i.e., x=2x=2 or x=10x=10)
  3. (x+6)2=16(x+6)^2=16, so x=6±4x=-6 \pm 4 (i.e., x=2x=-2 or x=10x=-10) (correct answer)
  4. (x+12)2=124(x+12)^2=124, so x=12±124x=-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 ax2+bx+c=0ax^2 + bx + c = 0 into the form (xp)2=q(x - p)^2 = q, which is easy to solve by taking square roots: xp=±qx - p = \pm \sqrt{q}, giving x=p±qx = 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±iqx = p \pm i \sqrt{|q|}. The value of q immediately reveals the nature of solutions. To solve x2+12x+20=0x^2 + 12x + 20 = 0: (1) Move constant: x2+12x=20x^2 + 12x = -20. (2) Half of 12 is 6, square 36, add both sides: x2+12x+36=20+36=16x^2 + 12x + 36 = -20 + 36 = 16. (3) (x+6)2=16(x + 6)^2 = 16. (4) x+6=±4x + 6 = \pm 4. (5) x=6±4x = -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(b/2)^2, (4) Add to BOTH sides, (5) Factor left as (x+b/2)2(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!