Quadratics
The discriminant and the nature of roots
What you need to be able to do
- 1.1Find the discriminant \(b^2-4ac\) of a quadratic and use it to decide how many real roots the equation \(ax^2+bx+c=0\) has.
- 1.1Know the language: two distinct real roots, equal (repeated) roots, real roots, no real roots.
- 1.1Find the set of values of a constant \(k\) for which an equation has a given number of roots; show that an expression is always positive.
Before you start: Solving quadratic equations. Next: Quadratic inequalities (you need them to finish most discriminant questions).
Where the discriminant comes from, what it tells you and how to use it, with our own examples. Past-paper questions come in Part 2.
What you should already know
Reading a, b and c
Put the equation in the form \(ax^2+bx+c=0\) first, then read the coefficients with their signs. In \(4x^2-x-7=0\): \(a=4,\ b=-1,\ c=-7\). A coefficient can contain a letter: in \(x^2+kx+(k+3)=0\), \(b=k\) and \(c=k+3\).
Quick check: \(a,b,c\) for \(3x^2=5-2x\)?
\(3x^2+2x-5=0\): \(a=3,\ b=2,\ c=-5\)
The quadratic formula
\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). It comes from completing the square (previous page).
Quick check: solve \(x^2-6x+7=0\)
\(x=\tfrac{6\pm\sqrt8}2=3\pm\sqrt2\)
Square roots of negatives
No real number squared is negative, so \(\sqrt{-5}\) is not a real number. \(\sqrt0=0\).
Quick check: which of \(\sqrt{9},\ \sqrt0,\ \sqrt{-9}\) are real?
\(\sqrt9=3\) and \(\sqrt0=0\); \(\sqrt{-9}\) is not real
Squaring an expression
Square the whole thing: \((3k)^2=9k^2\), \((k-2)^2=k^2-4k+4\), \((-k)^2=k^2\).
Quick check: expand \((2k+1)^2\)
\(4k^2+4k+1\)
Where the discriminant comes from
Look at the three equations below. They differ only in the constant, yet one has two roots, one has a single root and one has none. The formula shows why.
| Equation | \(b^2-4ac\) | Formula gives | Roots |
|---|---|---|---|
| \(x^2-6x+5=0\) | \(36-20=16\) | \(\tfrac{6\pm\sqrt{16}}2=\tfrac{6\pm4}2\) | \(x=5,\ 1\): two different roots |
| \(x^2-6x+9=0\) | \(36-36=0\) | \(\tfrac{6\pm\sqrt0}2=\tfrac{6\pm0}2\) | \(x=3\) twice: one repeated root |
| \(x^2-6x+13=0\) | \(36-52=-16\) | \(\tfrac{6\pm\sqrt{-16}}2\) | no real roots |
The \(\pm\) part, \(\pm\sqrt{b^2-4ac}\), is what splits the answer into two. If the number under the root is positive, \(+\) and \(-\) give two different roots. If it is zero, \(+0\) and \(-0\) are the same, so the two roots coincide at \(x=-\tfrac b{2a}\). If it is negative, there is no real square root at all. So the number \(b^2-4ac\) discriminates between the three cases: that is its name.
The picture. Completing the square on \(ax^2+bx+c\) gives \(a\left(x+\tfrac b{2a}\right)^2-\tfrac{b^2-4ac}{4a}\). So for a ∪-shaped curve (\(a>0\)) the vertex is at height \(-\tfrac{b^2-4ac}{4a}\): below the axis when the discriminant is positive (two crossings), on the axis when it is zero (it touches), and above the axis when it is negative (it misses).
What the discriminant tells you
In the formula \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), everything depends on the number under the square root. That number is the discriminant.
| Discriminant | Roots of ax² + bx + c = 0 | Graph of y = ax² + bx + c | Words used in questions |
|---|---|---|---|
| \(b^2-4ac>0\) | two different real roots (\(\pm\) a positive root) | crosses the x-axis twice | "two distinct real roots", "two distinct points" |
| \(b^2-4ac=0\) | one root, \(x=-\tfrac{b}{2a}\), counted twice | touches the x-axis (vertex on the axis) | "equal roots", "repeated root", "exactly one root", "tangent" |
| \(b^2-4ac<0\) | no real roots (square root of a negative) | misses the x-axis completely | "no real roots", "does not meet", "no solutions" |
| \(b^2-4ac\ge0\) | at least one real root | meets the x-axis | "real roots", "meets", "intersects" |
- How many real roots does \(2x^2+3x+5=0\) have?
Answer
\(9-40=-31<0\): none - How many real roots does \(4x^2-12x+9=0\) have?
Answer
\(144-144=0\): one repeated root, \(x=\tfrac32\) - How many real roots does \(x^2+x-3=0\) have?
Answer
\(1+12=13>0\): two distinct real roots
Finding the values of k: the method
- Rearrange to \(ax^2+bx+c=0\) (everything on one side) and read off a, b and c with any k in them. Write them down: \(a=2,\ b=3k,\ c=k\). A coefficient can be an expression such as \(k+8\).
- Choose the condition from the table: \(>0\), \(=0\), \(<0\) or \(\ge0\).
- Substitute into \(b^2-4ac\) with brackets: \((3k)^2-4(2)(k)\). Write this line out on its own.
- Simplify to a quadratic (or linear) expression in \(k\).
- Solve: for \(=0\), solve the equation; for \(>\) or \(<\), find the critical values then the region, with a quick sketch.
- Write the set with the right signs: strict \(<\), \(>\) unless the condition includes \(=\).
Step 5 is a quadratic inequality in \(k\): the method is on the next page, Quadratic inequalities. The short version: sketch \(y=(k-p)(k-q)\), a ∪ shape; \(>0\) is outside the roots, \(<0\) is between them.
Case by case
1 No real roots, k only in b
\(5x^2+kx+5=0\) has no real roots. Find the set of values of \(k\).
Askedset of kKnowa = 5, b = k, c = 5Routeb² − 4ac < 0, solve, inside region
- \(k^2-4(5)(5)<0\Rightarrow k^2<100\)
- Critical values \(k=\pm10\); \(k^2-100<0\) between them
- \(\mathbf{-10<k<10}\) (also accepted: \(|k|<10\), \((-10,10)\))
Three common mistakes: keeping only the positive value (k < 10), choosing the outside regions instead of the inside one, and stopping at the critical values ±10.
2 Distinct roots, k in b and c
\(3x^2+2kx+k=0\) has distinct real roots. Find the set of values of \(k\).
Askedset of kKnowa = 3, b = 2k, c = kRouteb² − 4ac > 0, factorise, outside regions
- \((2k)^2-4(3)(k)>0\Rightarrow4k^2-12k>0\)
- \(4k(k-3)>0\): critical values \(0\) and \(3\)
- \(\mathbf{k<0,\ k>3}\)
3 k in the x² coefficient as well
\(2kx^2+(k+6)x+4=0\) has two distinct real roots. Find the set of values of \(k\).
Askedset of kKnowa = 2k, b = k + 6, c = 4Routebracket the whole b, expand carefully
- \((k+6)^2-4(2k)(4)>0\Rightarrow k^2+12k+36-32k>0\)
- \(k^2-20k+36>0\Rightarrow(k-2)(k-18)>0\)
- \(\mathbf{k<2,\ k>18}\) (strict: at \(k=2\) or \(18\) the roots are equal, not distinct)
Square the whole of b: (k + 6)², not k² + 6. (Strictly, k = 0 makes the equation linear; schemes have not asked you to remove it.)
4 Exactly one root (repeated)
\(9x^2-12x+7=k\) has exactly one root. Find \(k\) and the root.
Askedk and the rootKnowmove k across: c = 7 − kRouteb² − 4ac = 0 gives k; the root is −b/2a
- \(9x^2-12x+(7-k)=0\): \(12^2-4(9)(7-k)=0\Rightarrow7-k=4\Rightarrow k=3\)
- Repeated root \(x=-\tfrac{b}{2a}=\tfrac{12}{18}=\mathbf{\tfrac23}\)
With the completed square, 9x² − 12x + 7 = (3x − 2)² + 3, you can read k = 3 and x = 2/3 straight off.
5 Always positive
Show that \(x^2-4x+7>0\) for all \(x\).
Askeda proofKnowa = 1 > 0Route∪ shape that never meets the axis: D < 0, or complete the square
- \(b^2-4ac=16-28=-12<0\): no real roots
- \(a=1>0\), so the curve is ∪ and lies wholly above the \(x\)-axis
- Therefore \(x^2-4x+7>0\) for all \(x\). (Or: \((x-2)^2+3\ge3>0\).)
D < 0 alone isn't enough: x² − 4x + 7 and −x² + 4x − 7 both have D < 0. You also need a > 0.
- Find the values of \(k\) for which \(x^2+kx+9=0\) has equal roots.
Answer
\(k^2-36=0\): \(k=\pm6\) - Find the set of values of \(k\) for which \(kx^2+4x+1=0\) has no real roots.
Answer
\(16-4k<0\): \(k>4\) - Find the set of values of \(k\) for which \(x^2+(k-2)x+1=0\) has two distinct real roots.
Answer
\((k-2)^2-4>0\Rightarrow k(k-4)>0\): \(k<0,\ k>4\)
Key facts at a glance
- Discriminant \(=b^2-4ac\): the number under the root in the formula.
- \(>0\) two distinct real roots; \(=0\) one repeated root at \(x=-\tfrac b{2a}\); \(<0\) no real roots; \(\ge0\) real roots.
- Graph: crosses, touches or misses the \(x\)-axis.
- With a letter \(k\): write \(a, b, c\) in terms of \(k\), substitute with brackets, solve the equation or inequality in \(k\).
- Always positive: \(a>0\) and \(b^2-4ac<0\).
How the discriminant is examined, what the mark scheme wants, then real Paper 1 questions worked through slowly.
How this is examined
About ten Paper 1 questions since 2020 ask for the values of \(k\) that give no, equal or two distinct roots (2-5 marks), and many more hide the same step inside a line-and-curve question. The marks follow the method: M1 for using \(b^2-4ac\) with the right condition (the scheme wants it seen "not just in the quadratic formula"), A1 for the simplified expression in \(k\), DM1 for solving it with a method shown, A1 for the final set.
The final mark is lost most often on the answer form. The November 2022 report listed the usual errors: "only considering positive values of k", "deciding that the outside regions were the correct answer", or giving only the critical values. Schemes give B0 for \(\le\) or \(\ge\) when the roots must be distinct.
Mark-scheme focus
Find the set of values of k for which … has no real roots
![(a) Find the set of values of k for which the equation 8x² + kx + 2 = 0 has no real roots. [2]](/notes-content/9709/quadratics/img/9709_w22_12_q03_a.png)
- \(k^2-4\times8\times2<0\): use of \(b^2-4ac\), not just inside the formula M1
- \(-8<k<8\) A1; also \(|k|<8\) or \((-8,8)\)
- \(k<8\) only (forgetting the negative root)
- \(k<-8,\ k>8\) (the outside regions: that is where \(b^2-4ac>0\))
- \(k=\pm8\), or \(\sqrt{64}\) left unsimplified
"No real roots" means \(b^2-4ac<0\): an inside region. The November 2022 report: candidates often considered only positive values of k or chose the outside regions.
"\(-8<k\) or \(k<8\)" and "\(-8<k\) and \(k<8\)" were condoned on this scheme; newer schemes do not allow "or" for an inside region, so write \(-8<k<8\).
From: 9709/12/O/N/22 Q3(a) and the November 2022 report.
Use the quadratic formula to show that … has no solutions
![The function f is given by f(x) = 4cos⁴x + cos²x − k for 0 ≤ x ≤ 2π. (a) Given k = 3, find the exact solutions of f(x) = 0. [5] (b) Use the quadratic formula to show that, when k > 5, the equation f(x) = 0 has no solutions. [5]](/notes-content/9709/quadratics/img/9709_s22_12_q11.png)
- \(\cos^2x=\dfrac{-1\pm\sqrt{1+16k}}{8}\) with the formula used *M1
- the minus root is negative, so impossible for \(\cos^2x\) B1
- for \(k>5\): \(\sqrt{1+16k}>9\), so the plus root \(>1\), impossible DM1 A1
- a concluding statement covering both cases A1
- one numerical value of \(k\) (e.g. \(k=6\)) - that doesn't prove it for all \(k>5\)
- ignoring the instruction to use the formula
- considering only one of the two roots
This is not b² − 4ac < 0: the discriminant 1 + 16k is positive. There ARE real values of cos²x; they just lie outside 0 ≤ cos²x ≤ 1. The June 2022 report calls this the hardest part of the paper.
From: 9709/12/M/J/22 Q11(b) and the June 2022 report.
- Set of k for no / equal / distinct roots: most papers, 2-5 marks, usually as a line-and-curve question (next pages).
- Discriminant inside another topic: trig equations with k 9709/12/F/M/20 Q11, tangents to circles, "a single point".
- Answer form: the final A1/B1 is for the correctly written set - see Quadratic inequalities.
Worked examples
Example 1 · k in all three coefficients (Paper 1, 4 marks) ★★ Medium
![Find the set of values of the constant k for which the quadratic equation 3kx² + (k + 8)x + 3 = 0 has two distinct real roots. [4]](/notes-content/9709/quadratics/img/9709_w25_11_q01.png)
A quadratic whose coefficients depend on \(k\). For some \(k\) it cuts the \(x\)-axis twice; for others it doesn't. We want the \(k\) values that give two different roots.
- two distinct real roots\(b^2-4ac>0\) (strictly)
- the constant kk appears in a and in b
- the set of values of \(k\)
- \(a=3k\)
- \(b=k+8\)bracket it
- \(c=3\)
- the answer is a range: an inequality (or two)
- Findset of k
- Use\((k+8)^2-4(3k)(3)>0\)
- Then\(k^2-20k+64>0\), factorise
- Region∪ in \(k\): \(>0\) is outside the roots
- Trap\(\le\)/\(\ge\), or the inside region
- Use\((k+8)^2-4(3k)(3)>0\)
Expand (k + 8)².
\(k^2+16k+64\).
Which two numbers multiply to 64 and add to −20?
\(-4\) and \(-16\): \((k-4)(k-16)\).
5 Now solve it have a go on paper first, then open the worked solution
- \(b^2-4ac=(k+8)^2-4(3k)(3)\) *M1
- \(=k^2-20k+64\) A1
- \(k^2-20k+64>0\Rightarrow(k-4)(k-16)>0\); critical values 4, 16 DM1
- \(\mathbf{k<4,\ k>16}\) B1
Test \(k=20\) (outside): \(D=400-400+64=64>0\) ✓. Test \(k=10\) (inside): \(D=100-200+64=-36<0\) ✓, correctly excluded.
Example 2 · Completed square or discriminant (Paper 1, 3 marks) ★ Easy
![(a) Express 4x² − 24x + p in the form a(x + b)² + c, where a and b are integers and c is to be given in terms of the constant p. [2] (b) Hence or otherwise find the set of values of p for which the equation 4x² − 24x + p = 0 has no real roots. [1]](/notes-content/9709/quadratics/img/9709_s23_12_q03.png)
Part (a) completes the square with \(p\) left as a letter. Part (b) asks when the quadratic never reaches zero. "Hence or otherwise" means either route is fine; the completed square is one line.
- Hence or otherwiseuse (a), or the discriminant
- no real rootsthe curve stays above the x-axis
- (a)\(a,\ b,\ c\) (\(c\) in terms of \(p\))
- (b)set of \(p\)
- \(4x^2-24x+p\)
- a = 4 > 0, so the least value is c
- (a)\(a(x+b)^2+c\)
- Use\(4(x^2-6x)+p=4(x-3)^2-36+p\)
- (b)no real roots
- Route 1least value \(p-36>0\)
- Route 2\(24^2-4(4)p<0\)
Why is p − 36 > 0 the condition?
\(4(x-3)^2\ge0\), so the least value is \(p-36\). If that is above 0, the expression is never 0.
Do both routes agree?
\(576-16p<0\Rightarrow p>36\). Yes.
5 Now solve it have a go on paper first, then open the worked solution
- (a) \(4(x-3)^2+p-36\): \(a=4,\ b=-3\) B1, \(c=p-36\) B1
- (b) \(p-36>0\) (or \(576-16p<0\)) \(\Rightarrow\mathbf{p>36}\) B1
The June 2023 report: "only a few saw the link with part (a), and instead usually restarted using the discriminant... simplifying the resulting inequality correctly proved challenging for many."
\(p=37\): \(4(x-3)^2+1>0\) ✓, never zero. \(p=36\): a repeated root at \(x=3\), so 36 is excluded ✓.
Example 3 · A trig equation that is a quadratic in tan x (Paper 1, 9 marks) ★★★ Hard
![(a) Solve the equation 3tan²x − 5tanx − 2 = 0 for 0° ≤ x ≤ 180°. [4] (b) Find the set of values of k for which the equation 3tan²x − 5tanx + k = 0 has no solutions. [2] (c) For the equation 3tan²x − 5tanx + k = 0, state the value of k for which there are three solutions in the interval 0° ≤ x ≤ 180°, and find these solutions. [3]](/notes-content/9709/quadratics/img/9709_m20_12_q11.png)
Why it is hard: part (c) needs you to combine the discriminant idea with the graph of tan x; many found only k.
Treat \(\tan x\) as the unknown \(u\): \(3u^2-5u+k=0\). Every real value of \(u\) gives one angle in \(0°\)–\(180°\) - except \(u=0\), which gives two (0° and 180°). That is the key to part (c).
- no solutionsno real values of \(\tan x\): \(b^2-4ac<0\)
- three solutionstwo values of \(\tan x\), one of them 0 (which gives 0° and 180°)
- 0° ≤ x ≤ 180°tan has period 180°: one angle per value, except tan x = 0
- (a)x for k = −2
- (b)set of k
- (c)k and three solutions
- \(3\tan^2x-5\tan x+k=0\)
- \(0°\le x\le180°\)
- \(\tan x\) can take any real value, so only \(b^2-4ac\) limits it
- (b)no solutions
- Use\(25-4(3)k<0\Rightarrow k>\tfrac{25}{12}\)
- (c)three solutions
- Useone root must be \(\tan x=0\) ⇒ \(k=0\)
- Then\(3\tan^2x-5\tan x=0\Rightarrow\tan x=0\) or \(\tfrac53\)
- Useone root must be \(\tan x=0\) ⇒ \(k=0\)
How many x in 0°–180° satisfy tan x = 2?
One: 63.4°. (The next one, 243.4°, is outside.)
How many satisfy tan x = 0?
Two: 0° and 180°, both inside the closed interval.
5 Now solve it have a go on paper first, then open the worked solution
- (a) \((\tan x-2)(3\tan x+1)=0\) M1; \(\tan x=2\) or \(-\tfrac13\) A1; \(x=63.4°\), \(161.6°\) B1FT B1FT
- (b) No real \(\tan x\): \(25-4(3)(k)<0\) M1 \(\Rightarrow\mathbf{k>\tfrac{25}{12}}\) A1
- (c) \(k=0\) M1; \(\tan x(3\tan x-5)=0\Rightarrow\tan x=0\) or \(\tfrac53\) A1; \(x=0°,\ 180°,\ 59.0°\) A1
The March 2020 report: in (c) "those candidates with a good understanding of the tangent graph realised that an extra solution in the range could only come from the solution of tan x = 0."
\(\tan59.0°\approx1.664\approx\tfrac53\) ✓. For \(k=\tfrac{25}{12}\) there is exactly one value of tan x (repeated), so one solution: consistent with \(k>\tfrac{25}{12}\) giving none.
Summary
- \(b^2-4ac>0\): two distinct real roots; \(=0\): one repeated root; \(<0\): none; \(\ge0\): real roots.
- Write \(a,\ b,\ c\) with brackets (they may contain \(k\)); square the whole of \(b\).
- Solve the inequality in \(k\) with a sketch; strict signs unless "=" is allowed.
- "Always positive" needs \(a>0\) and \(b^2-4ac<0\) (or a completed square).
| If you see… | condition |
|---|---|
| two distinct real roots / two distinct points | \(b^2-4ac>0\) |
| equal roots / repeated root / exactly one root / tangent / touches | \(b^2-4ac=0\) |
| no real roots / does not meet / no solutions | \(b^2-4ac<0\) |
| real roots / meets / intersects | \(b^2-4ac\ge0\) |
| positive for all x | \(a>0\) and \(b^2-4ac<0\) |
Can I do this?
- I know the four discriminant conditions and the words that go with each.
- I can read off a, b, c with k in them and substitute into b² − 4ac with brackets.
- I can find the set of values of k for no / equal / distinct roots, including k in the x² coefficient.
- I can prove an expression is positive for all x (a > 0 and b² − 4ac < 0).
- I write the final set with the correct strict or non-strict signs.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
The multiple-choice and guided questions need JavaScript. The concept checks and exam practice below work without it.
Concept checks
Answer each one in your head first, then open it.
1Two students disagree: one says \(x^2+1=0\) has no roots, the other says no real roots. Who is right?Show answer
The second. \(b^2-4ac=-4<0\) means no real roots; Paper 1 only deals with real numbers, so always say "real".
2Why does "tangent" mean discriminant = 0?Show answer
A tangent meets the curve at exactly one point, so the equation for the meeting points has a repeated root.
3If \(a\) and \(c\) have opposite signs, why must there be two distinct real roots?Show answer
Then \(-4ac>0\), so \(b^2-4ac>0\) whatever \(b\) is.
Exam-style practice
New questions, not used above. Do them on paper, then mark your work against the official mark scheme.
![Find the set of values of m for which the line with equation y = mx + 1 and the curve with equation y = 3x² + 2x + 4 intersect at two distinct points. [4]](/notes-content/9709/quadratics/img/9709_s20_13_q01.png)
Show the mark scheme
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(3x^2+2x+4=mx+1\Rightarrow3x^2+(2-m)x+3=0\) | B1 | one side equal to 0 | |
| \((2-m)^2-36\) | M1 | discriminant seen or implied | |
| \((m+4)(m-8)\) or \(2-m=\pm6\) | A1 | critical values \(-4\) and \(8\) | |
| \(m<-4,\ m>8\) | A1 | outside regions, strict signs | |
| Total: 4 marks | |||
Examiner tip Also on the scheme: \(\dfrac{dy}{dx}=6x+2=m\) at the points of contact gives the critical values. Either way, "two distinct points" means \(b^2-4ac>0\).
![A curve has equation y = x² + 2cx + 4 and a straight line has equation y = 4x + c, where c is a constant. Find the set of values of c for which the curve and line intersect at two distinct points. [5]](/notes-content/9709/quadratics/img/9709_m22_12_q02.png)
Show the mark scheme
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x^2+2cx+4=4x+c\Rightarrow x^2+(2c-4)x+4-c=0\) | *M1 | equate the y's, everything on one side | |
| \((2c-4)^2-4(4-c)\) | DM1 | discriminant with their coefficients | |
| \(=4c^2-12c\) | A1 | ||
| \(4c(c-3)>0\) | M1 | '> 0' applied, both regions considered | |
| \(c<0,\ c>3\) | A1 | must be in terms of \(c\); \(c\le0,\ c\ge3\) gets SC B1 only | |
| Total: 5 marks | |||
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