Unit 7 · Algebra · 4 teaching hours · 18 min read

Quadratic Equation

What the syllabus asks for 4

Nature of roots
Relation between roots and coefficients
Formation of quadratic equation
Symmetric roots; one or both roots common

A quadratic equation is any equation that can be written ax^{2} + bx + c = 0 with a \ne 0. You have solved these since Grade 9. What is new here is that you will answer questions about the roots without finding them: how many real roots there are, what they add up to, what they multiply to, and how to build an equation backwards from its roots. One expression — the discriminant — controls almost all of it.

The two facts this whole unit runs on

For ax^{2}+bx+c = 0 with roots \alpha and \beta,

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}, \qquad D = b^{2}-4ac

Almost every question is solved by writing down the sum, the product and the discriminant before doing anything else.

7.1 Nature of roots

Completing the square on ax^{2}+bx+c = 0 gives the quadratic formula.

The quadratic formula

Divide by a (allowed, since a \ne 0) and move the constant across:

x^{2} + \frac{b}{a}x = -\frac{c}{a}

Add \left(\dfrac{b}{2a}\right)^{2} to both sides to complete the square:

x^{2} + \frac{b}{a}x + \frac{b^{2}}{4a^{2}} = \frac{b^{2}}{4a^{2}} - \frac{c}{a} = \frac{b^{2}-4ac}{4a^{2}}

\left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2}-4ac}{4a^{2}}

Taking square roots and rearranging,

x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Everything about the kind of roots sits under the square-root sign.

Discriminant

D = b^{2} - 4ac

is the discriminant of ax^{2}+bx+c = 0. It discriminates — it decides — between the possible kinds of roots.

Assume a, b, c are real.

Discriminant	Nature of the roots	The graph y = ax^{2}+bx+c
D > 0, a perfect square	real, distinct, rational	cuts the x-axis at two rational points
D > 0, not a perfect square	real, distinct, irrational (a conjugate surd pair)	cuts the x-axis at two points
D = 0	real and equal, both = -b/2a	touches the x-axis at one point
D < 0	no real roots; a conjugate complex pair	does not meet the x-axis

The last two rows of the table assume a, b, c are rational; irrational and complex roots of a rational quadratic always come in conjugate pairs, so you can never have exactly one irrational root.

**Figure 1.** The three cases. D>0 gives two crossings, D=0 gives one touch at the vertex, D<0 gives no crossing at all.

"No roots" is wrong

When D < 0 the equation still has two roots — they are just not real. Write "the roots are imaginary (complex conjugates)", never "there are no roots".

Worked example 7.1

Problem. Discuss the nature of the roots of 2x^{2} - 7x + 3 = 0, then solve it.

Solution. Here a = 2, b = -7, c = 3.

D = b^{2} - 4ac = (-7)^{2} - 4(2)(3) = 49 - 24 = 25

D > 0 and 25 = 5^{2} is a perfect square, so the roots are real, distinct and rational.

x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4} = 3 \ \text{ or } \ \frac{1}{2}

Worked example 7.2

Problem. Find the values of k for which x^{2} + 2(k+1)x + 9k - 5 = 0 has equal roots.

Solution. Equal roots means D = 0. With a = 1, b = 2(k+1), c = 9k-5,

D = 4(k+1)^{2} - 4(9k-5) = 0 \;\Longrightarrow\; (k+1)^{2} - (9k-5) = 0

k^{2} + 2k + 1 - 9k + 5 = 0 \;\Longrightarrow\; k^{2} - 7k + 6 = 0

(k-1)(k-6) = 0 \;\Longrightarrow\; k = 1 \ \text{ or } \ k = 6

Check k = 1: the equation is x^{2}+4x+4 = 0, i.e. (x+2)^{2} = 0 — equal roots. Check k = 6: x^{2}+14x+49 = 0, i.e. (x+7)^{2} = 0. Both work.

Worked example 7.3

Problem. For what value of m does (m+1)x^{2} + 2(m+3)x + (m+8) = 0 have equal roots?

Solution. Set D = 0:

4(m+3)^{2} - 4(m+1)(m+8) = 0 \;\Longrightarrow\; (m+3)^{2} = (m+1)(m+8)

m^{2} + 6m + 9 = m^{2} + 9m + 8 \;\Longrightarrow\; -3m + 1 = 0 \;\Longrightarrow\; m = \frac{1}{3}

Note m = \tfrac13 keeps m+1 \ne 0, so the equation really is quadratic. (Had the answer been m = -1 we would have had to reject it.)

Always check that a \ne 0

In a question with a parameter in the x^{2} coefficient, a value that makes that coefficient zero must be rejected — the equation stops being quadratic.

7.2 Relation between roots and coefficients

Sum and product of the roots

If \alpha and \beta are the roots of ax^{2}+bx+c = 0, then ax^{2}+bx+c factorises as a(x-\alpha)(x-\beta). Expanding,

a(x-\alpha)(x-\beta) = ax^{2} - a(\alpha+\beta)x + a\alpha\beta

Comparing coefficients with ax^{2}+bx+c:

-a(\alpha+\beta) = b \;\Longrightarrow\; \alpha+\beta = -\frac{b}{a}, \qquad a\alpha\beta = c \;\Longrightarrow\; \alpha\beta = \frac{c}{a}

From these two, every symmetric expression in \alpha and \beta can be found without ever solving the equation. Write s = \alpha+\beta and p = \alpha\beta.

The standard identities

\alpha^{2}+\beta^{2} = s^{2} - 2p, \qquad (\alpha-\beta)^{2} = s^{2} - 4p

\alpha^{3}+\beta^{3} = s^{3} - 3ps, \qquad \frac{1}{\alpha}+\frac{1}{\beta} = \frac{s}{p}, \qquad \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{s^{2}-2p}{p}

Note that (\alpha-\beta)^{2} = s^{2}-4p = \dfrac{b^{2}-4ac}{a^{2}} = \dfrac{D}{a^{2}} — the discriminant appears again, which is why D = 0 means the roots coincide.

The sum and product also read off the graph: the axis of symmetry of y = ax^{2}+bx+c sits at x = -b/2a, exactly halfway between the roots.

**Figure 2.** The roots of x^{2}-5x+4 are 1 and 4. Their mean \\frac{5}{2} is the axis of symmetry x = -b/2a, and the vertex lies on it.

Worked example 7.4

Problem. If \alpha and \beta are the roots of 2x^{2}-5x+3 = 0, find (a) \alpha^{2}+\beta^{2}, (b) \alpha^{3}+\beta^{3}, (c) \dfrac{1}{\alpha}+\dfrac{1}{\beta}, (d) \alpha-\beta — all without solving.

Solution. s = \alpha+\beta = -\dfrac{-5}{2} = \dfrac52 and p = \alpha\beta = \dfrac32.

(a) \;\alpha^{2}+\beta^{2} = s^{2}-2p = \dfrac{25}{4} - 3 = \dfrac{13}{4}

(b) \;\alpha^{3}+\beta^{3} = s^{3} - 3ps = \dfrac{125}{8} - 3\left(\dfrac32\right)\left(\dfrac52\right) = \dfrac{125}{8} - \dfrac{45}{4} = \dfrac{125-90}{8} = \dfrac{35}{8}

(d) \;(\alpha-\beta)^{2} = s^{2}-4p = \dfrac{25}{4} - 6 = \dfrac14, so \alpha-\beta = \pm\dfrac12

Check. The roots really are \tfrac32 and 1; their sum is \tfrac52, their difference \tfrac12, and \left(\tfrac32\right)^{2}+1^{2} = \tfrac{13}{4}.

Worked example 7.5

Problem. If one root of x^{2}+px+q = 0 is double the other, show that 2p^{2} = 9q.

Solution. Let the roots be \alpha and 2\alpha.

\text{Sum: } \alpha + 2\alpha = 3\alpha = -p \;\Longrightarrow\; \alpha = -\frac{p}{3}

\text{Product: } \alpha(2\alpha) = 2\alpha^{2} = q

Substituting the first into the second,

2\left(-\frac{p}{3}\right)^{2} = q \;\Longrightarrow\; \frac{2p^{2}}{9} = q \;\Longrightarrow\; 2p^{2} = 9q

7.3 Formation of quadratic equation

Reverse the relation. If a quadratic with leading coefficient 1 has roots \alpha and \beta, then

(x-\alpha)(x-\beta) = 0 \;\Longrightarrow\; x^{2} - (\alpha+\beta)x + \alpha\beta = 0

Forming an equation

x^{2} - (\text{sum of roots})\,x + (\text{product of roots}) = 0

Find the sum and the product of the required roots, substitute, and clear fractions at the end. Never find the actual roots unless you have to.

Worked example 7.6

Problem. Form the quadratic equation whose roots are 2+\sqrt{3} and 2-\sqrt{3}.

Solution.

\text{Sum} = (2+\sqrt{3})+(2-\sqrt{3}) = 4

\text{Product} = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1

x^{2} - 4x + 1 = 0

Notice the surds cancel — that is the point of conjugate pairs, and it is why an equation with rational coefficients cannot have just one surd root.

Worked example 7.7

Problem. If \alpha, \beta are the roots of x^{2}-6x+8 = 0, form the equation whose roots are \alpha^{2} and \beta^{2}.

Solution. s = 6, p = 8. For the new roots:

\text{Sum} = \alpha^{2}+\beta^{2} = s^{2}-2p = 36-16 = 20

\text{Product} = \alpha^{2}\beta^{2} = (\alpha\beta)^{2} = 64

x^{2} - 20x + 64 = 0

Check. The original roots are 2 and 4, so the new roots should be 4 and 16: 4+16 = 20 and 4 \times 16 = 64. Correct.

Worked example 7.8

Problem. If \alpha, \beta are the roots of x^{2}+px+q = 0, form the equation whose roots are \dfrac{\alpha}{\beta} and \dfrac{\beta}{\alpha}.

Solution. Here s = -p and \alpha\beta = q.

\text{Sum} = \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^{2}+\beta^{2}}{\alpha\beta} = \frac{s^{2}-2q}{q} = \frac{p^{2}-2q}{q}

\text{Product} = \frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha} = 1

x^{2} - \frac{p^{2}-2q}{q}x + 1 = 0 \;\Longrightarrow\; qx^{2} - \left(p^{2}-2q\right)x + q = 0

7.4 Symmetric roots; one or both roots common

Symmetric functions of the roots

An expression in \alpha and \beta is symmetric if swapping \alpha and \beta leaves it unchanged — for example \alpha^{2}+\beta^{2}, \alpha^{3}\beta^{3}, or \dfrac{1}{\alpha}+\dfrac{1}{\beta}. Every symmetric expression can be written in terms of s and p alone, so it can be evaluated straight from the coefficients. That is exactly what §7.2 and §7.3 exploit.

\alpha - \beta is not symmetric (swapping gives \beta-\alpha), which is why we can only ever pin down (\alpha-\beta)^{2} and are left with a \pm.

One root common

Suppose a_1x^{2}+b_1x+c_1 = 0 and a_2x^{2}+b_2x+c_2 = 0 share exactly one root \alpha. In an exam you usually have numbers, and the fastest route is direct:

The three-line method for a common root

Solve whichever equation factorises.
Substitute each of its roots into the other equation.
Whichever value of the unknown makes it true is the answer.

There is also a general condition. Substituting x = \alpha in both and eliminating \alpha between them (by cross-multiplication) gives

\left(c_1a_2 - c_2a_1\right)^{2} = \left(a_1b_2 - a_2b_1\right)\left(b_1c_2 - b_2c_1\right)

and the common root itself is

\alpha = \frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}

Both roots common

If both roots are common, the two quadratics are the same equation up to a constant multiple, so their coefficients are proportional:

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

**Figure 3.** y = x^{2}-5x+6 and y = x^{2}-3x+2 share the root x = 2 only: both curves cross the x-axis at the same point, but their other roots differ.

Worked example 7.9

Problem. Find the value of k for which x^{2}-5x+6 = 0 and x^{2}-3x+k = 0 have a common root.

Solution. The first equation factorises: (x-2)(x-3) = 0, so its roots are 2 and 3. One of these must satisfy the second equation.

If the common root is 2: \;4 - 6 + k = 0 \Rightarrow k = 2.

If the common root is 3: \;9 - 9 + k = 0 \Rightarrow k = 0.

So k = 2 or k = 0.

Check k = 2: x^{2}-3x+2 = (x-1)(x-2), roots 1 and 2 — shares 2. Check k = 0: x^{2}-3x = x(x-3), roots 0 and 3 — shares 3. Both valid.

Worked example 7.10

Problem. The equations 2x^{2}+3x+5 = 0 and ax^{2}+bx+c = 0 have both roots in common. If a = 6, find b and c.

Solution. Both roots common means the coefficients are proportional:

\frac{a}{2} = \frac{b}{3} = \frac{c}{5}

With a = 6 the common ratio is \dfrac{6}{2} = 3, so

b = 3 \times 3 = 9, \qquad c = 3 \times 5 = 15

The second equation is 6x^{2}+9x+15 = 0, which is just 3 times the first — so of course the roots are identical.

Worked example 7.11

Problem. Find the value of k for which the roots of kx^{2} - 2(1+2k)x + 3(2+k) = 0 are equal.

Solution. Set D = 0 with a = k, b = -2(1+2k), c = 3(2+k):

4(1+2k)^{2} - 4k\cdot 3(2+k) = 0 \;\Longrightarrow\; (1+2k)^{2} - 3k(2+k) = 0

1 + 4k + 4k^{2} - 6k - 3k^{2} = 0 \;\Longrightarrow\; k^{2} - 2k + 1 = 0

(k-1)^{2} = 0 \;\Longrightarrow\; k = 1

Since k = 1 \ne 0, the equation is genuinely quadratic. Check: at k=1 the equation is x^{2}-6x+9 = 0, i.e. (x-3)^{2} = 0 — equal roots x = 3.

Worked example 7.12

Problem. If the equations x^{2}+bx+c = 0 and x^{2}+cx+b = 0 (b \ne c) have a common root, find that root and the relation between b and c.

Solution. Let \alpha be the common root. Then

\alpha^{2}+b\alpha+c = 0 \qquad \text{and} \qquad \alpha^{2}+c\alpha+b = 0

Subtracting the second from the first removes \alpha^{2}:

(b-c)\alpha + (c-b) = 0 \;\Longrightarrow\; (b-c)(\alpha - 1) = 0

Since b \ne c we may divide by b-c, giving \alpha = 1.

Putting \alpha = 1 back into the first equation:

1 + b + c = 0 \;\Longrightarrow\; b + c = -1

So the common root is 1 and b+c+1 = 0. (Subtracting the two equations is the standard first move whenever both are monic — it turns a pair of quadratics into one linear equation.)

Chapter summary

ax^{2}+bx+c = 0 has roots x = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.
Discriminant D = b^{2}-4ac: D > 0 real distinct (D a perfect square makes them rational), D = 0 real and equal, D < 0 a conjugate complex pair.
\alpha+\beta = -b/a and \alpha\beta = c/a; and (\alpha-\beta)^{2} = D/a^{2}.
Useful identities: \alpha^{2}+\beta^{2} = s^{2}-2p, \alpha^{3}+\beta^{3} = s^{3}-3ps, \dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{s}{p}.
To form an equation: x^{2} - (\text{sum})x + (\text{product}) = 0.
Symmetric expressions in the roots can always be written in s and p; \alpha-\beta cannot, so it carries a \pm.
One root common: substitute the known roots of one equation into the other; in general (c_1a_2-c_2a_1)^{2} = (a_1b_2-a_2b_1)(b_1c_2-b_2c_1).
Both roots common: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.

Practice questions

Group A — Multiple choice (1 mark each)

The roots of x^{2}-5x+6 = 0 are [1]a-2, -3b2, 3c1, 6d-1, -6
The roots of ax^{2}+bx+c = 0 are real and equal if [1]ab^{2}-4ac > 0bb^{2}-4ac = 0cb^{2}-4ac < 0db^{2} = 4a
If \alpha, \beta are the roots of x^{2}-3x+2 = 0, then \alpha+\beta is [1]a-3b2c3d-2
The quadratic equation whose roots are 3 and -2 is [1]ax^{2}-x-6 = 0bx^{2}+x-6 = 0cx^{2}-x+6 = 0dx^{2}+5x+6 = 0
If the roots of ax^{2}+bx+c = 0 are reciprocals of each other, then [1]aa = bbb = cca = cda+b+c = 0
If D < 0 for a quadratic with real coefficients, the graph of y = ax^{2}+bx+c [1]acuts the x-axis twicebtouches the x-axiscnever meets the x-axisdis a straight line

Answers to Group A

1. (b) — (x-2)(x-3) = 0.

2. (b) — D = 0 makes the \pm\sqrt{D} term vanish.

3. (c) — \alpha+\beta = -b/a = 3.

4. (a) — sum = 1, product = -6, so x^{2}-x-6 = 0.

5. (c) — if the roots are \alpha and 1/\alpha then the product c/a = 1, so a = c.

6. (c) — no real root means no real x with y = 0.

Group B — Short answer (5 marks each)

Discuss the nature of the roots of 3x^{2}-2x+1 = 0 and of 4x^{2}-12x+9 = 0. [5]
Find the values of k for which x^{2} - 2(1+3k)x + 7(3+2k) = 0 has equal roots. [5]
If \alpha, \beta are the roots of 2x^{2}+3x-4 = 0, find \alpha^{2}+\beta^{2} and \dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}. [5]
If \alpha, \beta are the roots of 3x^{2}-5x+2 = 0, form the equation whose roots are \dfrac{1}{\alpha} and \dfrac{1}{\beta}. [5]
Find the value of k if x^{2}-7x+12 = 0 and x^{2}-8x+k = 0 have a common root. [5]
If one root of x^{2}+px+q = 0 is three times the other, show that 3p^{2} = 16q. [5]

Answers to Group B

1. For 3x^{2}-2x+1 = 0: D = (-2)^{2}-4(3)(1) = 4-12 = -8 < 0, so the roots are imaginary, a conjugate complex pair.

For 4x^{2}-12x+9 = 0: D = 144 - 4(4)(9) = 144-144 = 0, so the roots are real and equal, both x = -b/2a = 12/8 = \tfrac32. (Indeed 4x^{2}-12x+9 = (2x-3)^{2}.)

2. D = 0 gives 4(1+3k)^{2} - 28(3+2k) = 0, i.e. (1+3k)^{2} - 7(3+2k) = 0.

1 + 6k + 9k^{2} - 21 - 14k = 0 \;\Longrightarrow\; 9k^{2} - 8k - 20 = 0

k = \frac{8 \pm \sqrt{64 + 720}}{18} = \frac{8 \pm 28}{18} = 2 \ \text{ or } \ -\frac{10}{9}

3. s = -\tfrac32, p = -2.

\alpha^{2}+\beta^{2} = s^{2}-2p = \frac94 + 4 = \frac{25}{4}

\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} = \frac{\alpha^{2}+\beta^{2}}{(\alpha\beta)^{2}} = \frac{25/4}{4} = \frac{25}{16}

4. For 3x^{2}-5x+2 = 0, s = \tfrac53 and p = \tfrac23. The new roots have

\text{sum} = \frac{1}{\alpha}+\frac{1}{\beta} = \frac{s}{p} = \frac{5/3}{2/3} = \frac52, \qquad \text{product} = \frac{1}{\alpha\beta} = \frac{1}{2/3} = \frac32

x^{2} - \frac52 x + \frac32 = 0 \;\Longrightarrow\; 2x^{2}-5x+3 = 0

(Sensible: reversing the coefficients of a quadratic inverts its roots. The original roots are \tfrac23 and 1; the new ones are \tfrac32 and 1.)

5. x^{2}-7x+12 = (x-3)(x-4), so its roots are 3 and 4.

If the common root is 3: 9 - 24 + k = 0 \Rightarrow k = 15. If the common root is 4: 16 - 32 + k = 0 \Rightarrow k = 16.

So k = 15 or k = 16.

6. Let the roots be \alpha and 3\alpha. Then 4\alpha = -p, so \alpha = -p/4; and 3\alpha^{2} = q. Substituting,

3\left(\frac{p^{2}}{16}\right) = q \;\Longrightarrow\; 3p^{2} = 16q

Group C — Long answer (8 marks each)

(a) Derive the quadratic formula by completing the square. [4](b) Define the discriminant and describe the nature of the roots in each of its three cases, giving one numerical example of each. [4]
(a) If \alpha, \beta are the roots of ax^{2}+bx+c = 0, prove that \alpha+\beta = -\dfrac{b}{a} and \alpha\beta = \dfrac{c}{a}. [3](b) If \alpha, \beta are the roots of x^{2}-6x+8 = 0, form the equation whose roots are \alpha+\dfrac{1}{\beta} and \beta+\dfrac{1}{\alpha}. [5]
(a) Show that if the two equations a_1x^{2}+b_1x+c_1 = 0 and a_2x^{2}+b_2x+c_2 = 0 have both roots common then \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}. [4](b) Find k so that x^{2}-5x+6=0 and x^{2}-3x+k=0 have a common root, and in each case state the common root. [4]

Answers to Group C

1. (a) See the derivation in §7.1: divide by a, complete the square by adding (b/2a)^{2}, and take square roots to get x = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.

(b) D = b^{2}-4ac. D > 0: real and distinct — e.g. x^{2}-5x+6 = 0 with D = 1, roots 2, 3. D = 0: real and equal — e.g. x^{2}-4x+4 = 0, double root 2. D < 0: a conjugate complex pair — e.g. x^{2}+x+1 = 0 with D = -3.

2. (a) Since \alpha, \beta are the roots, ax^{2}+bx+c \equiv a(x-\alpha)(x-\beta) = ax^{2} - a(\alpha+\beta)x + a\alpha\beta. Comparing the coefficients of x and the constants gives \alpha+\beta = -b/a and \alpha\beta = c/a.

(b) For x^{2}-6x+8=0: s = 6, p = 8 (the roots are 2 and 4).

\text{Sum} = \alpha + \beta + \frac{1}{\alpha} + \frac{1}{\beta} = s + \frac{s}{p} = 6 + \frac68 = \frac{27}{4}

\text{Product} = \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = 8 + 2 + \frac18 = \frac{81}{8}

x^{2} - \frac{27}{4}x + \frac{81}{8} = 0 \;\Longrightarrow\; 8x^{2} - 54x + 81 = 0

Check. The new roots are 2+\tfrac14 = \tfrac94 and 4+\tfrac12 = \tfrac92; their sum is \tfrac{27}{4} and product \tfrac{81}{8}. Correct.

3. (a) Let the common roots be \alpha and \beta. Then a_1x^{2}+b_1x+c_1 = a_1(x-\alpha)(x-\beta) and a_2x^{2}+b_2x+c_2 = a_2(x-\alpha)(x-\beta). Dividing one identity by the other,

\frac{a_1x^{2}+b_1x+c_1}{a_2x^{2}+b_2x+c_2} = \frac{a_1}{a_2}

for all x, so the two quadratics are proportional term by term: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.

(b) Roots of the first are 2 and 3. Putting x = 2 in the second: 4-6+k = 0, so k = 2 (common root 2). Putting x = 3: 9-9+k = 0, so k = 0 (common root 3).

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