5.3 Quadratic Formula and Completing the Square

5.3 Quadratic Formula and Completing the Square

When factoring is slow or the integers refuse to cooperate, AAF still expects the real solutions of ax^2 + bx + c = 0. The quadratic formula always applies (provided a ≠ 0):

x = (−b ± √(b^2 − 4ac)) / (2a)

Memorize it as a single fraction. A common error writes −b ± √(b^2 − 4ac) / 2a without grouping the denominator, which looks as if only the radical is divided by 2a. For a = 2 and −b = 3, that slip turns (3 ± 5)/4 into 3 ± 5/4.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Discriminant: two real, one real, no real

AAF focuses on real solutions. Complex numbers are not the emphasis of this placement test.

You can decide the number of real roots without computing the roots. For 2x^2 − 3x − 2, D = 9 − 4(2)(−2) = 9 + 16 = 25 > 0, so two distinct reals. For x^2 + 4x + 8, D = 16 − 32 = −16 < 0, so no real solution. For x^2 − 6x + 9, D = 36 − 36 = 0, so one real root x = 3.

Watch the sign of c inside −4ac. If c is already negative, −4ac is positive when a is positive: that is how 2x^2 − 3x − 2 gained +16 rather than −16.

Worked example: 2x^2 − 3x − 2 = 0

a = 2, b = −3, c = −2.

D = (−3)^2 − 4(2)(−2) = 9 + 16 = 25.

√25 = 5.

x = (3 ± 5) / 4

First root: (3 + 5)/4 = 8/4 = 2.

Second root: (3 − 5)/4 = (−2)/4 = −1/2.

So x = 2 or x = −1/2.

These factor as well: 2x^2 − 3x − 2 = (2x + 1)(x − 2). The formula and factoring agree. On the CAT, if you see integer or fraction roots, either method is valid; if you see unsimplified radicals in the options, the formula was the intended path. Notice −b is +3 because b was −3. Dropping that sign is the fastest way to land on (−3 ± 5)/4.

Simplify radicals in exact answers

Leave answers in exact form unless the stem asks for a decimal.

x^2 + 4x − 7 = 0

x = (−4 ± √(16 + 28)) / 2 = (−4 ± √44) / 2 = (−4 ± 2√11) / 2 = −2 ± √11

Simplify √44 = √(4 · 11) = 2√11 before dividing by 2. Options may include (−4 ± √44)/2, −2 ± √11, and a decimal. The first two are equivalent; AAF often wants the fully simplified exact expression −2 ± √11.

Another: 3x^2 − 2x − 2 = 0

D = 4 + 24 = 28 = 4 · 7

x = (2 ± 2√7) / 6 = (1 ± √7) / 3

Cancel a common factor of 2 from 2 ± 2√7 and 6; do not cancel the 2 into the 3 only on one term.

When a calculator icon appears, decimal approximations are tempting: √11 ≈ 3.317, so −2 + √11 ≈ 1.317. AAF frequently lists exact equivalents instead. Compute the decimal only to match an option that is already rounded; do not round your own answer into a nearby distractor such as 1.3 or 4/3.

Completing the square as a solving method

Goal: rewrite ax^2 + bx + c = 0 so one side is a perfect square.

When a = 1. Solve x^2 − 6x + 5 = 0 by completing the square (this is the same parabola as section 5.1).

x^2 − 6x = −5

Take half of −6, which is −3, square it: 9.

x^2 − 6x + 9 = −5 + 9

(x − 3)^2 = 4

x − 3 = ±2

x = 5 or x = 1

Those are the factored roots. Completing the square also produced vertex form y = (x − 3)^2 − 4 when you treat y instead of 0.

When a ≠ 1. First divide through by a, or factor a out of the x-terms.

Solve 2x^2 − 3x − 2 = 0 by completing the square.

2(x^2 − (3/2)x) = 2

Half of 3/2 is 3/4; square is 9/16.

2(x^2 − (3/2)x + 9/16) = 2 + 2 · (9/16)

2(x − 3/4)^2 = 2 + 9/8 = 25/8

(x − 3/4)^2 = 25/16

x − 3/4 = ±5/4

x = 3/4 + 5/4 = 2 or x = 3/4 − 5/4 = −1/2

Same two roots as the formula. The extra arithmetic is why the formula is popular; completing the square is still the way you derive the formula and the way you reach vertex form.

Completing the square as a path to vertex form

For graphing, you do not set y to 0. Start with y = 2x^2 − 8x + 5.

y = 2(x^2 − 4x) + 5

y = 2(x^2 − 4x + 4 − 4) + 5

y = 2((x − 2)^2 − 4) + 5

y = 2(x − 2)^2 − 8 + 5

y = 2(x − 2)^2 − 3

Vertex (2, −3), opens up, axis x = 2. The y-intercept is still (0, 5). This conversion does double duty on the Skills Insight ladder: the vertex it exposes is the 250–262 graph–algebra connection, and completing the square itself — solving a quadratic by any method — is a 276–300 technique.

Check with −b/(2a): h = 8/4 = 2, k = 2(4) − 8(2) + 5 = 8 − 16 + 5 = −3. Completing the square and the axis formula agree.

A second vertex conversion

y = −x^2 + 4x − 1

y = −(x^2 − 4x) − 1

y = −(x^2 − 4x + 4 − 4) − 1

y = −((x − 2)^2 − 4) − 1

y = −(x − 2)^2 + 4 − 1

y = −(x − 2)^2 + 3

Opens down, vertex (2, 3), range y ≤ 3. That range is a parabola statement, not a quartic statement. The functions chapter handles higher-degree range; here the completed square is enough.

Exact answers versus calculator decimals

College Board’s calculator policy: some AAF questions show an on-screen calculator icon at the top-right corner. Handheld calculators are not allowed except with an approved accommodation. When the icon is absent, exact arithmetic is expected. When the icon is present, you may still need √50 = 5√2 because the correct option is simplified radical form.

Simplify this fully: x = (6 ± √72) / 6.

√72 = √(36 · 2) = 6√2

x = (6 ± 6√2) / 6 = 1 ± √2

Options that stop at (6 ± 6√2)/6 are equivalent but not simplified. Options that compute √72 ≈ 8.49 and then (6 + 8.49)/6 ≈ 2.415 will not match 1 + √2 if you round too early. Match the option’s form: exact simplified, exact unsimplified, or rounded decimal — do not invent a fourth form.

Putting the formula on a created equation

A rectangle’s length is 5 cm more than its width, and the area is 84 cm².

w(w + 5) = 84

w^2 + 5w − 84 = 0

D = 25 + 336 = 361 = 19^2

w = (−5 ± 19) / 2

w = 7 or w = −12. Width 7 cm, length 12 cm. The negative root is algebraically valid for the equation and invalid as a length. AAF solve-the-equation items keep both; possible-width items keep 7.

Formula, factoring, and completing the square are three routes to the same real roots. Use the discriminant to know how many real answers to expect before you box them. Inequalities and line-and-parabola systems use those roots next, in Quadratic Inequalities and Quadratic-Linear Systems.