Complete the Square to Find Solutions - Algebra 2

A student attempts to complete the square for $4x^2 + 12x - 5 = 0$ and writes $4(x + \frac{3}{2})^2 = 14$ . To verify this is correct, which calculation should be performed?
Answer: Expand $4(x + \frac{3}{2})^2 - 14$ and check if it equals $4x^2 + 12x - 5$ When verifying algebraic manipulations like completing the square, you need to check that both forms of the equation are truly equivalent. The student transformed $4x^2 + 12x - 5 = 0$ into $4(x + \frac{3}{2})^2 = 14$ , which means they're claiming that $4x^2 + 12x - 5$ equals $4(x + \frac{3}{2})^2 - 14$ . To verify this claim, you should expand $4(x + \frac{3}{2})^2 - 14$ and check if it produces the original expression. Let's see: $4(x + \frac{3}{2})^2 - 14 = 4(x^2 + 3x + \frac{9}{4}) - 14 = 4x^2 + 12x + 9 - 14 = 4x^2 + 12x - 5$ . Perfect! This confirms the student's work is correct. Choice A suggests factoring, but $4x^2 + 12x - 5$ doesn't factor nicely, and this approach doesn't directly verify the algebraic manipulation. Choice B has the wrong sign—it asks you to expand $4(x + \frac{3}{2})^2 + 14$ , which would give $4x^2 + 12x + 23$ , not matching the original expression. Choice C involves solving and substituting, which is unnecessarily complicated and doesn't directly verify that the two algebraic forms are equivalent. Study tip: When checking completing-the-square work, always verify by expanding the completed form and ensuring it matches the original expression. The key is recognizing that $ax^2 + bx + c = 0$ becomes $a(x + h)^2 = k$ , so $ax^2 + bx + c$ should equal $a(x + h)^2 - k$ .
Solve using completing the square: $x^2+2x+5=0.$ (Your completed-square form should look like $(x-p)^2=q$ .)
Answer: $(x+1)^2=-4$ , so $x=-1 \pm 2i$ . 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 $ax^2 + bx + c = 0$ 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 $x^2 + 2x + 5 = 0$ 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!
Solve using the method of completing the square: $x^2+2x+5=0.$ Express solutions in $a \pm bi$ form.
Answer: $(x+1)^2=-4$ , so $x=-1 \pm 2i$ This question tests your ability to use completing the square to find complex solutions, expressing them in $a \pm bi$ form. 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 $x^2 + 2x + 5 = 0$ : (1) Move constant: $x^2 + 2x = -5$ . (2) Half of 2 is 1, square 1, add both sides: $x^2 + 2x + 1 = -5 + 1 = -4$ . (3) $(x + 1)^2 = -4$ . (4) $x + 1 = \pm 2i$ . (5) $x = -1 \pm 2i$ . Wonderful work with imaginaries! Choice A correctly completes the square to get the proper form and expresses the complex solutions accurately. Choice B treats $q$ as positive, ignoring the negative and missing the $i$ terms. 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 the correct answer every time! You're excelling at this!
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?
Answer: Both forms are equivalent and correct, but Student B's form is more useful for solving 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.
Complete the square on the general quadratic equation to derive the quadratic formula. Starting from $ax^2+bx+c=0 \quad (a\ne 0),$ which expression for $x$ results from completing the square?
Answer: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ 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 $ax^2 + bx + c = 0$ 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 derive the quadratic formula, complete the square on $ax^2 + bx + c = 0$ treating a, b, c as constants: divide by a to get $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ , then complete square on $x^2 + \frac{b}{a}x$ by adding $\left(\frac{b}{2a}\right)^2$ , which gives $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ , then take square root and solve to get $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Choice C correctly shows the quadratic formula with the proper negative sign on b and the discriminant $b^2 - 4ac$ . Choice A incorrectly has $b^2 + 4ac$ under the square root—the discriminant must be $b^2 - 4ac$ , not $b^2 + 4ac$ ! The systematic derivation by completing the square reveals why the quadratic formula has exactly this form, including the crucial minus sign in the discriminant.
Solve the quadratic equation by completing the square: $x^2+8x+7=0.$ Write your work in the form $(x-p)^2=q$ and then give the solutions for $x$ .
Answer: $(x+4)^2=9$ , so $x=-4\pm 3$ (i.e., $x=-1$ or $x=-7$ ) This question tests your ability to use completing the square to solve quadratic equations and understand how it leads to the solutions, showing it's a reliable method for any quadratic. 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 + 8x + 7 = 0 by completing the square: (1) Move constant to right: x² + 8x = -7. (2) Take half of 8 to get 4, square it to get 16, add to both sides: x² + 8x + 16 = -7 + 16 = 9. (3) Factor left side as perfect square: (x + 4)² = 9. (4) Take square roots: x + 4 = ±3. (5) Solve: x = -4 ± 3, so x = -1 or x = -7. The systematic process works every time! 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, but B uses something leading to 57, which 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 the correct answer every time! Keep practicing, you've got this!
Solve by completing the square and express the solutions in complex form (if needed): $x^2-6x+13=0$
Answer: $(x-3)^2=-4$ , so $x=3\pm 2i$ 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 $ax^2 + bx + c = 0$ 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.
Complete the square to solve and express any nonreal solutions in $a+bi$ form: $x^2-6x+13=0.$
Answer: $(x-3)^2=-4$ , so $x=3\pm 2i$ . 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 $ax^2 + bx + c = 0$ 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 $x^2 - 6x + 13 = 0$ by completing the square: (1) Move constant: $x^2 - 6x = -13$ . (2) Half of -6 is -3, square 9, add: $x^2 - 6x + 9 = -4$ . (3) $(x - 3)^2 = -4$ . (4) $x - 3 = \pm 2i$ . (5) $x = 3 \pm 2i$ . 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 gets positive $q=4$ , but actual $q=-4$ for complex roots. 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!
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?
Answer: $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ 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 $ax^2 + bx + c = 0$ 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!
The equation $x^2 + 6x + k = 0$ is completed to the square form $(x + 3)^2 = m$ . If this equation has exactly one solution, what is the relationship between $k$ and $m$ ?
Answer: $k = 9$ and $m = 0$ , giving one repeated solution When you see a quadratic equation that needs to be completed to the square and has "exactly one solution," you're dealing with a perfect square trinomial where the discriminant equals zero. Let's complete the square for $x^2 + 6x + k = 0$ . To transform this into $(x + 3)^2 = m$ , we take half of the coefficient of $x$ (which is 6), square it: $(\frac{6}{2})^2 = 9$ . So we rewrite as: $x^2 + 6x + 9 - 9 + k = 0$ $(x + 3)^2 - 9 + k = 0$ $(x + 3)^2 = 9 - k$ This means $m = 9 - k$ . For exactly one solution, the right side must equal zero (since $\sqrt{0} = 0$ gives only one value). Therefore: $9 - k = 0$ , so $k = 9$ and $m = 0$ . Choice A incorrectly sets $k = 0$ , which would give $m = 9 - 0 = 9$ , not $m = 9$ as stated. This would yield two solutions: $x = 0$ and $x = -6$ . Choice B uses $k = -9$ , giving $m = 9 - (-9) = 18$ , which creates two solutions since $\sqrt{18} = \pm 3\sqrt{2}$ . Choice C has $k = 3$ , so $m = 9 - 3 = 6$ , again yielding two solutions since $\sqrt{6} = \pm\sqrt{6}$ . Choice D correctly identifies $k = 9$ and $m = 0$ , giving the repeated solution $x = -3$ . Key takeaway: A quadratic has exactly one solution when it's a perfect square trinomial, which happens when the constant term makes the completed square form equal zero.
Transform and solve by completing the square: $x^2+10x-3=0.$ Write it as $(x-p)^2=q$ and then solve for $x$ .
Answer: $(x+5)^2=28$ , so $x=-5\pm \sqrt{28}=-5\pm 2\sqrt{7}$ . 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 $ax^2 + bx + c = 0$ 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 $x^2 + 10x - 3 = 0$ by completing the square: (1) Move constant: $x^2 + 10x = 3$ . (2) Half of 10 is 5, square 25, add: $x^2 + 10x + 25 = 28$ . (3) $(x + 5)^2 = 28$ . (4) $x + 5 = \pm \sqrt{28} = \pm 2\sqrt{7}$ . (5) $x = -5 \pm 2\sqrt{7}$ . 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 D miscalculates q as 22 instead of 28 after adding 25 to 3. 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!
Solve by completing the square: $x^2+12x+20=0$ Write it as $(x-p)^2=q$ and solve.
Answer: $(x+6)^2=16$ , so $x=-6\pm 4$ 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 $ax^2 + bx + c = 0$ 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 $x = -2$ or $x = -10$ . 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 $x = -2$ or $x = -10$ .
Solve the equation by completing the square: $x^2 + 4x + 13 = 0$ Express solutions in the form $a \pm bi$ .
Answer: $ $(x + 2)^2$ = -9 $, so $x = -2 \pm 3i$ 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 $ax^2 + bx + c = 0$ 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 $x^2 + 4x + 13 = 0$ : (1) Move constant: $x^2 + 4x = -13$ . (2) Take half of 4 to get 2, square it to get 4, add to both sides: $x^2 + 4x + 4 = -13 + 4 = -9$ . (3) Factor: $(x + 2)^2 = -9$ . (4) Take square roots: $x + 2 = \pm \sqrt{-9} = \pm 3i$ . (5) Solve: $x = -2 \pm 3i$ . Choice A correctly completes the square to get $(x + 2)^2 = -9$ and recognizes that $\sqrt{-9} = 3i$ , giving complex solutions. Choice C incorrectly calculates the right side as positive 9 instead of negative 9—remember $-13 + 4 = -9$ , not $+9$ ! The negative value on the right side after completing the square signals that you'll get complex solutions.
Use completing the square to solve: $3x^2+18x+15=0$ (First divide by the leading coefficient, then complete the square.)
Answer: $(x+3)^2=4$ , so $x=-3\pm 2$ 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 $a x^2 + b x + c = 0$ 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 $x = -1$ or $x = -5$ . 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.
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?
Answer: $c < 16$ and $q = 16 - c$ 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).
Solve using completing the square (do not factor): $x^2 + 12x + 20 = 0.$
Answer: $(x+6)^2=16$ , so $x=-6 \pm 4$ (i.e., $x=-2$ or $x=-10$ ) 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!
Solve by completing the square: $x^2-12x+20=0.$
Answer: $(x-6)^2=16$ , so $x=6\pm 4$ (i.e., $x=10$ or $x=2$ ). 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 $ax^2 + bx + c = 0$ 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 $x^2 - 12x + 20 = 0$ 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!
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?
Answer: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ 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!
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.
Answer: $(x+4)^2=9$ , so $x=-4 \pm 3$ (i.e., $x=-1$ or $x=-7$ ). 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 $ax^2 + bx + c = 0$ 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 $x = -1$ or $x = -7$ . 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!
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 $c$ ?
Answer: $q = \frac{b^2 - 4ac}{4a}$ 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.

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