8 min readPathway sheet2-mark and 3-mark questions

Solving by the quadratic formula

The roots of ax² + bx + c = 0 are x = (−b ± √(b² − 4ac))/(2a) when b² − 4ac ≥ 0. Use it when factorising is hard.

Quadratic formulax = (−b ± √(b² − 4ac))/(2a)

DiscriminantD = b² − 4ac

Real roots exist whenb² − 4ac ≥ 0

Method

1
Write the equation in standard form ax² + bx + c = 0.
2
Write the values of a, b and c, with their signs.
3
Find the discriminant D = b² − 4ac.
4
If D < 0, there are no real roots. Stop here.
5★
If D ≥ 0, substitute in x = (−b ± √D)/(2a).
6
Work out the root with + and the root with − separately, and simplify.

★ key step

Example 1: rational roots

Problem

Solve 3x² − 5x − 2 = 0 using the quadratic formula.

Working

a = 3, b = −5, c = −2
D = (−5)² − 4(3)(−2) = 25 + 24 = 49
x = (−(−5) ± √49)/(2 × 3) = (5 ± 7)/6
x = (5 + 7)/6 = 12/6 = 2, or x = (5 − 7)/6 = −2/6 = −1/3
Check x = 2: 3(4) − 10 − 2 = 0.

Answer

x = 2 or x = −1/3

Example 2: irrational roots

Problem

Solve x² − 4x + 1 = 0 using the quadratic formula.

Working

a = 1, b = −4, c = 1
D = 16 − 4 = 12, and √12 = √(4 × 3) = 2√3
x = (4 ± 2√3)/2 = 2 ± √3
Check: sum of roots = 4 = −b/a and product = 4 − 3 = 1 = c/a.

Answer

x = 2 + √3 or x = 2 − √3

Example 3: a speed problem

Problem

A cyclist rides 30 km at a uniform speed. If her speed were 5 km/h more, she would take 1 hour less. Find her speed.

Working

Let her speed be x km/h. Time = distance ÷ speed.
30/x − 30/(x + 5) = 1
Multiply by x(x + 5): 30(x + 5) − 30x = x(x + 5), so 150 = x² + 5x.
x² + 5x − 150 = 0, with a = 1, b = 5, c = −150.
D = 25 + 600 = 625, and √625 = 25.
x = (−5 ± 25)/2, so x = 10 or x = −15. Reject −15, since speed cannot be negative.
Check: 30/10 = 3 hours and 30/15 = 2 hours, which is 1 hour less.

Answer

Her speed is 10 km/h.

Common mistakes

Using b instead of −b: when b = −5, −b = +5.
Dividing only √D by 2a. The whole numerator −b ± √D is divided by 2a.
Squaring a negative b wrongly: (−5)² = 25, not −25.
Not simplifying surds: write √12 as 2√3 and √40 as 2√10.

In real life

Engineers use the quadratic formula to find when a ball, a rocket or a water jet reaches a given height. A shop owner can use it to find a price that gives a target income.

Exam angle

1 marks: “For 2x² − 3x − 5 = 0, find the value of b² − 4ac.” (−3)² − 4(2)(−5) = 9 + 40 = 49.

2 marks: “Find the roots of 2x² + x − 4 = 0 using the quadratic formula.” a = 2, b = 1, c = −4. D = 1 + 32 = 33. x = (−1 ± √33)/4.

3 marks: “Solve for x: x² − 2ax + a² − b² = 0.” D = 4a² − 4(a² − b²) = 4b², so √D = 2b. x = (2a ± 2b)/2, so x = a + b or x = a − b.

Remember it

"Minus b, plus or minus root of b squared minus 4ac, all over 2a." Say it aloud until it sounds like a song.