math-2017:A.REI.4math-2017:A.REI.4amath-2017:A.REI.4c
math-2017:A.REI.4 Solve quadratic equations in one variable.
math-2017:A.REI.4a Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x − p)² = q that has the same solutions.
math-2017:A.REI.4c Derive the quadratic formula using the method of completing the square.
Learning goal: Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x − p)² = q that has the same solutions, solve from that form, and derive the quadratic formula by completing the square on ax² + bx + c = 0.
Scope note. Part c of A.REI.4 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.
What a quadratic equation in one variable is. A quadratic equation in x is any equation that can be arranged as ax² + bx + c = 0 with a ≠ 0. The condition a ≠ 0 is part of the definition: if a were zero, the equation would no longer be quadratic: it would be linear if b ≠ 0, and if b = 0 as well it would be the constant statement c = 0, true for every x or for none. Solving it means finding every value of x that makes the equation true. A quadratic can have two real solutions, one repeated real solution, or no real solutions; the method in this lesson tells you which, and gives exact solutions with radicals rather than decimal guesses.
The idea behind completing the square. The expression x² + 6x is not a perfect square, but x² + 6x + 9 is, because (x + 3)² = x² + 6x + 9. The 9 is the square of half the coefficient of x: half of 6 is 3, and 3² = 9. In general, x² + bx + (b/2)² = (x + b/2)², because expanding (x + b/2)² gives x² + 2(b/2)x + (b/2)² = x² + bx + (b/2)². That identity is the whole method. If we can arrange an equation so that one side is x² + bx and then add (b/2)² to both sides, the left side becomes a perfect square and the equation takes the form (x − p)² = q. Adding the same number to both sides keeps the same solutions, so this transformation never creates or loses a solution. That is the reason the method is trustworthy: it is built entirely from reversible steps, unlike the squaring step in radical equations.
Solving from (x − p)² = q. If q > 0 there are two real solutions, x − p = √q and x − p = −√q, so x = p ± √q. If q = 0 there is one repeated solution, x = p. If q < 0 there are no real solutions, because a real number squared is never negative; in that case the solutions are complex, and the next lesson shows how to write them. Reading the sign of q is the fastest way to know how many real solutions a quadratic has.
A full model with leading coefficient 1. Solve x² + 6x − 7 = 0. Move the constant: x² + 6x = 7. Half of 6 is 3, and 3² = 9; add 9 to both sides: x² + 6x + 9 = 16, so (x + 3)² = 16. This is the form (x − p)² = q with p = −3 and q = 16. Take square roots: x + 3 = 4 or x + 3 = −4, so x = 1 or x = −7. Check both in the original equation: 1 + 6 − 7 = 0, and 49 − 42 − 7 = 0. The equation (x + 3)² = 16 and the equation x² + 6x − 7 = 0 have the same solutions, {1, −7}, which is what transforming an equation is supposed to guarantee.
When the leading coefficient is not 1. Divide every term by a first, so that the coefficient of x² is 1, then proceed. Solve 2x² − 12x + 5 = 0. Divide by 2: x² − 6x + 5/2 = 0. Move the constant: x² − 6x = −5/2. Half of −6 is −3, and (−3)² = 9; add 9 to both sides: (x − 3)² = 9 − 5/2 = 13/2. Here p = 3 and q = 13/2. Take square roots: x = 3 ± √(13/2). Rationalizing, √(13/2) = √26/2, so the exact solutions are x = 3 ± √26/2, approximately 5.5495 and 0.4505. Check the approximate one: 2(5.5495)² − 12(5.5495) + 5 ≈ 61.594 − 66.594 + 5 = 0.000, as close to zero as the rounding allows. The exact form is the answer; the decimal is the check and the estimate.
When the coefficient of x is odd. The method does not care. Solve x² + 3x − 1 = 0. Then x² + 3x = 1; half of 3 is 3/2, and (3/2)² = 9/4; so (x + 3/2)² = 1 + 9/4 = 13/4, and x = −3/2 ± √13/2 = (−3 ± √13)/2, approximately 0.3028 and −3.3028. Students who avoid fractions by trying to factor will find this one cannot be factored over the integers, and completing the square is what settles it.
Deriving the quadratic formula. Now do the same steps once, in general, on ax² + bx + c = 0 with a ≠ 0. Divide by a, which is allowed because a ≠ 0: x² + (b/a)x + c/a = 0. Move the constant: x² + (b/a)x = −c/a. Half of b/a is b/(2a), whose square is b²/(4a²); add it to both sides: x² + (b/a)x + b²/(4a²) = b²/(4a²) − c/a. The left side is now (x + b/(2a))². Put the right side over the common denominator 4a²: −c/a = −4ac/(4a²), so the right side is (b² − 4ac)/(4a²). The equation is (x + b/(2a))² = (b² − 4ac)/(4a²), which is the form (x − p)² = q with p = −b/(2a) and q = (b² − 4ac)/(4a²). Take square roots of both sides: x + b/(2a) = ±√(b² − 4ac)/(2a). The denominator is √(4a²) = 2|a|, but because the ± already allows both signs, writing 2a in place of 2|a| changes nothing. Subtract b/(2a): x = (−b ± √(b² − 4ac))/(2a). That is the quadratic formula, and every step was a step you have already performed on numbers. The quantity b² − 4ac under the root is the discriminant; its sign plays the role of the sign of q: positive gives two real solutions, zero gives one repeated solution, negative gives no real solutions.
Checking the formula against the method. For 2x² − 12x + 5 = 0, a = 2, b = −12, c = 5, so the formula gives x = (12 ± √(144 − 40))/4 = (12 ± √104)/4 = (12 ± 2√26)/4 = 3 ± √26/2, exactly what completing the square produced. The formula is not a separate fact to memorize; it is completing the square done once and for all.
The misconception and why it is wrong. The most common error is adding (b/2)² to the left side only, writing x² + 6x = 2 and then (x + 3)² = 2. This changes the equation: the left side gained 9 and the right side did not, so the new equation has different solutions, and none of them satisfy the original. The correct line is (x + 3)² = 11, with x = −3 ± √11 ≈ 0.3166 and −6.3166. A second error is dropping the ± and reporting only x = p + √q. A quadratic with q > 0 has two solutions, and the square root step must produce both.
Worked transformation and solution: Solve 3x² + 5x − 2 = 0 by completing the square. Divide by 3: x² + (5/3)x − 2/3 = 0, so x² + (5/3)x = 2/3. Half of 5/3 is 5/6, and (5/6)² = 25/36. Add it: (x + 5/6)² = 2/3 + 25/36 = 24/36 + 25/36 = 49/36. So p = −5/6 and q = 49/36. Take square roots: x + 5/6 = ±7/6, giving x = 2/6 = 1/3 or x = −12/6 = −2. Check x = 1/3: 3(1/9) + 5/3 − 2 = 1/3 + 5/3 − 6/3 = 0. Check x = −2: 12 − 10 − 2 = 0. The quadratic formula agrees: (−5 ± √(25 + 24))/6 = (−5 ± 7)/6.
Worked irrational solutions: Solve x² − 10x + 18 = 0. Then x² − 10x = −18; half of −10 is −5, squared is 25; so (x − 5)² = 25 − 18 = 7, and x = 5 ± √7, approximately 7.6458 and 2.3542. Exact check with x = 5 + √7: (5 + √7)² − 10(5 + √7) + 18 = 25 + 10√7 + 7 − 50 − 10√7 + 18 = 0. The √7 terms cancel, which is why the check works exactly and not just approximately.
Worked derivation used on a specific equation in parallel: Solve 4x² + 8x − 5 = 0 and, beside each step, write the general step. Divide by a = 4: x² + 2x − 5/4 = 0 (general: x² + (b/a)x + c/a = 0). Move the constant: x² + 2x = 5/4 (general: x² + (b/a)x = −c/a). Add (b/(2a))² = 1: (x + 1)² = 9/4 (general: (x + b/(2a))² = (b² − 4ac)/(4a²); here (64 + 80)/64 = 144/64 = 9/4, which matches). Roots: x + 1 = ±3/2 (general: x + b/(2a) = ±√(b² − 4ac)/(2a); here ±12/8 = ±3/2). Solutions: x = 1/2 or x = −5/2. Check: 4(1/4) + 4 − 5 = 0 and 4(25/4) − 20 − 5 = 25 − 25 = 0.
Worked reading of q: For x² + 5x + 7 = 0, completing the square gives (x + 5/2)² = 25/4 − 7 = 25/4 − 28/4 = −3/4. Because q = −3/4 is negative, no real number x satisfies the equation; the discriminant confirms it, 25 − 28 = −3 < 0. The solutions are complex and are written in the next lesson.
Method record: For each of x² − 8x + 3 = 0, 2x² + 4x − 3 = 0 and x² + 7x + 2 = 0, write a two-column record: the left column carries every step of completing the square with the numbers, the right column carries the same step written for ax² + bx + c = 0. End each record with the equation in the form (x − p)² = q, the values of p and q, the exact solutions, decimal approximations to four places, and the substitution check of one exact solution showing the radical terms cancelling.
Independent performance: Produce a written derivation of the quadratic formula from ax² + bx + c = 0, with a ≠ 0, in which every line is justified by naming the operation (divide both sides by a, subtract c/a from both sides, add (b/(2a))² to both sides, rewrite the left side as a perfect square, combine the right side over 4a², take square roots of both sides, subtract b/(2a)). Then choose three quadratic equations of your own, one with two rational solutions, one with two irrational solutions and one with no real solutions, and solve each twice: by completing the square to the form (x − p)² = q and by the formula. Show that both routes give the same solutions and substitute each real solution back. Success criteria: the derivation states where a ≠ 0 is used and where the ± enters; each transformed equation is shown to have the same solutions as the original because only reversible operations were used; and every real solution is checked in the original equation.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Transform x² − 8x + 3 = 0 into the form (x − p)² = q, state p and q, and solve.
Move the constant: x² − 8x = −3. Half of −8 is −4, squared is 16; add 16 to both sides: (x − 4)² = 13. So p = 4 and q = 13. Then x = 4 ± √13, approximately 7.6056 and 0.3944. Check x = 4 + √13: (4 + √13)² − 8(4 + √13) + 3 = 16 + 8√13 + 13 − 32 − 8√13 + 3 = 0. Adding 16 to both sides is reversible, so the transformed equation has the same solutions as the original.
Why is the number added when completing the square always the square of half the coefficient of x?
Because (x + b/2)² expands to x² + 2(b/2)x + (b/2)² = x² + bx + (b/2)². So x² + bx becomes a perfect square exactly when (b/2)² is added; any other constant leaves an expression that is not the square of a binomial. The constant is determined by b alone, which is why the method works for any quadratic once the leading coefficient is 1.
Solve 4x² + 8x − 5 = 0 by completing the square and check both solutions.
Divide by 4: x² + 2x = 5/4. Add 1: (x + 1)² = 9/4, so x + 1 = ±3/2 and x = 1/2 or x = −5/2. Check x = 1/2: 4(1/4) + 4 − 5 = 1 + 4 − 5 = 0. Check x = −5/2: 4(25/4) − 20 − 5 = 25 − 25 = 0. Dividing by 4 first is the convenient route; adding 16 to 4x² + 8x would not produce a perfect square in x (another correct route adds 4 to both sides of 4x² + 8x = 5 to get (2x + 2)² = 9).
In the derivation of the quadratic formula, identify the step that requires a ≠ 0 and the step at which the ± sign appears.
Dividing every term by a to get x² + (b/a)x + c/a = 0 requires a ≠ 0, since division by zero is undefined (and if a = 0 the equation is not quadratic). The ± appears when taking square roots of both sides of (x + b/(2a))² = (b² − 4ac)/(4a²): a positive number has two square roots, one positive and one negative, so x + b/(2a) = ±√(b² − 4ac)/(2a). Dropping either sign loses a solution.
Complete the square on x² + 5x + 7 = 0 and explain what the result says about real solutions.
x² + 5x = −7; half of 5 is 5/2, squared is 25/4; (x + 5/2)² = 25/4 − 7 = −3/4. So q = −3/4 < 0. The square of a real number cannot be negative, so there is no real solution. The discriminant 25 − 28 = −3 says the same thing; the solutions are complex conjugates.
Use the quadratic formula on 2x² + 3x − 9 = 0 and confirm by substitution.
a = 2, b = 3, c = −9. Discriminant 9 − 4(2)(−9) = 9 + 72 = 81, whose square root is 9. x = (−3 ± 9)/4, so x = 6/4 = 3/2 or x = −12/4 = −3. Check x = 3/2: 2(9/4) + 9/2 − 9 = 9/2 + 9/2 − 9 = 0. Check x = −3: 18 − 9 − 9 = 0. The discriminant 81 is a perfect square, so with these integer coefficients both solutions are rational, and the equation factors as (2x − 3)(x + 3). A whole-number discriminant that is not a perfect square, such as 8 for x² − 2 = 0, gives irrational solutions.
A student solving x² + 6x = 2 writes (x + 3)² = 2. Find the error, correct it and solve.
The student added 9 to the left side to complete the square but not to the right side, so the new equation is not equivalent to the old one. Correct line: (x + 3)² = 2 + 9 = 11, so x = −3 ± √11, approximately 0.3166 and −6.3166. Check x = −3 + √11: (−3 + √11)² + 6(−3 + √11) = 9 − 6√11 + 11 − 18 + 6√11 = 2, as required.