A. Quadratic Functions · Quadratic Expressions and Equations

Using the Quadratic Formula

A1.6About 70 minutes

Learning goals

I can rewrite an equation in the form ax² + bx + c = 0 and identify a, b and c.
I can use the quadratic formula with a calculator to find exact and approximate roots.
I can solve real problems with the formula and check my roots with graphing technology.

Words to know

quadratic formula
x = (−b ± √(b² − 4ac)) / (2a), the roots of ax² + bx + c = 0, a ≠ 0.

discriminant
The expression b² − 4ac under the square root.

exact root
A root written with radicals or fractions, such as (5 + √37)/6.

approximate root
A rounded decimal value of a root, such as 1.85.

graphing technology
A graphing calculator or software (such as Desmos or GeoGebra) that draws graphs and finds intercepts.

One formula, every quadratic equation

In the last lesson you saw where the quadratic formula comes from. Now you will use it, quickly and accurately, on equations that would be hard or impossible to factor, including the decimal coefficients that real situations produce. The curriculum asks you to solve these using technology: a scientific calculator for the arithmetic, and graphing technology (a graphing calculator, Desmos or GeoGebra) to check your roots.

The parts of the quadratic formula. The expression under the root, b² − 4ac, is called the discriminant; it gets its own lesson: Real Roots, x-Intercepts and the Discriminant.

Using the formula: four steps

Rearrange to ax² + bx + c = 0.
Write down a, b and c, with their signs.
Compute the discriminant b² − 4ac first (use brackets for negatives: (−5)² = 25).
Substitute into x = (−b ± √(b² − 4ac)) / (2a). Work out the + root and the − root separately.

Calculator tip: store or write down √(b² − 4ac) to several decimal places, then compute each root. Round only the final answers (usually to two decimal places).

Worked example

Example 1: A trinomial that does not factor

Solve 3x² − 5x − 1 = 0. Round to two decimal places.

a = 3, b = −5, c = −1.
Discriminant: (−5)² − 4(3)(−1) = 25 + 12 = 37.
x = (5 ± √37) / 6. √37 ≈ 6.0828.
x ≈ (5 + 6.0828) / 6 ≈ 1.85 and x ≈ (5 − 6.0828) / 6 ≈ −0.18.
Check x ≈ 1.847: 3(3.412) − 5(1.847) − 1 ≈ 10.24 − 9.24 − 1 = 0 ✓ (rounding makes it close to 0, not exactly 0).

x ≈ 1.85 or x ≈ −0.18 (exactly, x = (5 ± √37)/6)

Worked example

Example 2: Rearrange first

Solve 2x(x − 3) = 7.

Expand and rearrange: 2x² − 6x − 7 = 0. So a = 2, b = −6, c = −7.
Discriminant: 36 − 4(2)(−7) = 36 + 56 = 92. √92 ≈ 9.5917.
x = (6 ± 9.5917) / 4.
x ≈ 15.5917 / 4 ≈ 3.90 or x ≈ −3.5917 / 4 ≈ −0.90.

x ≈ 3.90 or x ≈ −0.90

Worked example

Example 3: A thrown ball

A ball is thrown upward from 1.5 m above the ground. Its height in metres after t seconds is h = −4.9t² + 20t + 1.5. (a) When is the ball 15 m high? (b) When does it land?

(a) Set h = 15: −4.9t² + 20t + 1.5 = 15, so −4.9t² + 20t − 13.5 = 0. a = −4.9, b = 20, c = −13.5.
Discriminant: 400 − 4(−4.9)(−13.5) = 400 − 264.6 = 135.4. √135.4 ≈ 11.636.
t = (−20 ± 11.636) / (−9.8): t ≈ (−8.364) / (−9.8) ≈ 0.85 or t ≈ (−31.636) / (−9.8) ≈ 3.23. Both make sense: once going up, once coming down.
(b) Set h = 0: −4.9t² + 20t + 1.5 = 0. Discriminant: 400 + 29.4 = 429.4. √429.4 ≈ 20.722.
t = (−20 ± 20.722) / (−9.8): t ≈ −0.07 or t ≈ 4.16. Reject the negative time (before the throw).

(a) at about 0.85 s and 3.23 s (b) it lands after about 4.16 s

Graphing technology confirms Example 3: the line h = 15 meets the parabola at t ≈ 0.85 s and t ≈ 3.23 s, and the ball reaches the ground at t ≈ 4.16 s.

Worked example

Example 4: The frame again

For the 30 cm by 20 cm photo, the frame's area is 4x² + 100x (x is the width in cm). What width makes the frame's area half the photo's area, 300 cm²?

4x² + 100x = 300, so 4x² + 100x − 300 = 0. Divide by 4: x² + 25x − 75 = 0.
Product −75 and sum 25: no integers work, so use the formula. Discriminant: 625 + 300 = 925, √925 ≈ 30.414.
x = (−25 ± 30.414) / 2: x ≈ 2.71 or x ≈ −27.71. Reject the negative width.
Check: 4(2.707)² + 100(2.707) ≈ 29.3 + 270.7 = 300 ✓.

The frame should be about 2.71 cm wide.

Avoid these slips

b² is never negative: (−6)² = +36. Type it as (−6)² with brackets, or just think "b times b".
−4ac: watch the signs. If a and c have opposite signs, −4ac is positive.
Divide the whole numerator by 2a. If a is negative, 2a is negative too.
Do not round √(b² − 4ac) to a whole number before finishing.

Just for fun

A calculator will happily tell you a ball lands at t = −0.07 s. Mathematics is honest; it just isn't always sensible. That part is your job.

Hands-on activity

Time a real toss

You need: a soft ball, a stopwatch app on a phone, and a measuring tape. Work outside or in a room with a high ceiling, with nothing breakable nearby; a grown-up or friend times you.

Hold the ball at about 1.5 m. Toss it straight up and let it fall to the ground while your helper times the whole flight. Repeat three times and average.
A good model for the height is h = −4.9t² + v·t + 1.5, where v is the launch speed in m/s. Use your average time T: the ball lands when h = 0, so v = (4.9T² − 1.5) ÷ T.
Using your v, solve −4.9t² + vt + 1.5 = 0 with the quadratic formula. Does the positive root match your measured time?
Use your model to predict the maximum height (hint: the vertex is halfway between the two roots).

What to look for: for example, a measured time of 1.6 s gives v = (4.9(2.56) − 1.5) ÷ 1.6 ≈ 6.9 m/s, and the formula's positive root is about 1.6 s, as it should be. The negative root is a time before the toss and is rejected. Small timing errors are normal; the model ignores air resistance.

Your turn

Solve x² + 3x − 5 = 0.

Solve 2x² − 7x + 2 = 0.

Solve 5x² + 2x − 1 = 0.

Solve x² = 4x + 7.

Solve 0.5x² + 1.2x − 3 = 0.

Solve 6x² + x − 2 = 0 with the formula. What does the perfect-square discriminant tell you?

Solve −2x² + 3x + 4 = 0.

More practice: warm-up

Name a, b and c in 3x² − x + 7 = 0.

Rewrite 4 − x² = 2x in the form ax² + bx + c = 0 and name a, b, c.

Find b² − 4ac for x² + 5x + 2 = 0.

What is (−5)²? What is −5²?

Solve x² − 2x − 1 = 0.

More practice: on your own

Solve 3x² + 4x − 2 = 0.

Solve (x + 2)(x − 1) = 5.

A ball's height is h = −4.9t² + 15t + 2 (metres, seconds). When does it land?

A rectangle's length is 4 m more than its width, and its area is 50 m². Find its dimensions to two decimal places.

Ava solved x² − 6x + 4 = 0 and wrote b² − 4ac = −36 − 16. Correct her work.

Challenge

For the 30 cm by 20 cm photo, what frame width gives a frame area of 200 cm²? (The frame area is 4x² + 100x.)

The canoe outfitter's revenue is R = −10x² + 90x + 3600, where x is the number of $2 increases. For which whole numbers x is the revenue at least $3700?

Solve x² − 4x + 1 = 0 and give the roots in exact simplified form.

Check your understanding

Quiz

For 2x² − 3x − 4 = 0, what are a, b and c?

For x² − 6x + 2 = 0, b² − 4ac equals:

Solve x² + 2x − 4 = 0 (to two decimal places).

A ball's landing time comes out as t ≈ 2.9 s or t ≈ −0.2 s. The answer is:

Which error gives a wrong answer with the formula?

What is the first step before using the quadratic formula on 3x² = 5x + 2?

For parents and teachers

Big idea (A1.6, second part): apply the quadratic formula, with a scientific calculator, to equations that do not factor, including real problems with decimal coefficients, and check with graphing technology.

How to help: ask your teen to talk you through the four steps aloud for one problem, including the signs of a, b and c. If they have a graphing app, ask them to show you that the x-intercepts match their answers.

Common difficulties:

Sign errors with b² and −4ac (typing −6² instead of (−6)² on a calculator gives −36).
Dividing only the square root by 2a.
Rounding too early, which can change the second decimal place.
Not rejecting impossible answers (negative times or lengths) in context.