What is 2.3: Quadratic functions?

For ax^2+bx+c, the discriminant b^2-4ac is positive for two distinct real roots, zero for a repeated root and negative for no real roots.

How many marks are 2.3 questions worth?

2.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.3 appeared in?

We have not yet indexed a verified past-paper appearance for 2.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.3

Work with quadratic functions and their graphs; the discriminant, including conditions for real and repeated roots; completing the square; solution of quadratic equations, including solving quadratic equations in a function of the unknown.

Draft — not yet indexed

Quadratic functions

Worked answers and methods for 2.3 on Edexcel A-level Maths 9MA0.

Explanation

For $ax^2+bx+c$ , the discriminant $b^2-4ac$ is positive for two distinct real roots, zero for a repeated root and negative for no real roots.
Complete the square to expose the turning point and range, or use factorisation and the quadratic formula to solve equations.
For example, $x^2-6x+11=(x-3)^2+2$ , so its minimum point is $(3,2)$ and it has no real roots.
When the quadratic is in a function of the unknown, substitute a new variable, solve the quadratic in that variable, then solve every resulting equation and reject invalid roots.
The substituted function may be a power, trigonometric function, exponential or logarithm, so enforce its range and domain when returning to $x$ .

Worked example

Find the two values of $k$ for which $2x^2+(k-1)x+8=0$ has a repeated root.

1.A repeated root requires discriminant zero: $(k-1)^2-4(2)(8)=0$ .
2.Hence $(k-1)^2=64$ , so $k-1=\pm8$ and therefore $k=9$ or $k=-7$ .

Answer: $k=9$ or $k=-7$

Common mistakes

Don't use $b^2-4ac=0$ for two distinct roots instead of for one repeated root.
Don't solve the quadratic in that variable and fail to return to solve for the original unknown after substituting a new variable.

Exam tip

For a repeated-root condition, write $b^2-4ac=0$ before substituting the coefficients.

Worked practice

Tier 1 · Easy

Write

x^2-8x+5

in completed-square form.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$(x-4)^2-11$	2
Notes
Half the coefficient of $x$ is $-4$ . Since $(x-4)^2=x^2-8x+16$ , subtract $11$ to restore the constant $5$ : $x^2-8x+5=(x-4)^2-11$ .

(2 marks)

Tier 2 · Standard

The curve with equation

y=x^2-6x+k

and the line with equation

y=2x-3

meet at exactly one point. Find the value of

k

and the coordinates of this point.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$k=13$ The point is $(4,5)$ .	4
Notes
At an intersection, $x^2-6x+k=2x-3$ , so $x^2-8x+k+3=0$ . Exactly one intersection requires a repeated root, hence $(-8)^2-4(k+3)=0$ . Therefore $64-4k-12=0$ and $k=13$ . The repeated root is $x=4$ , and the line gives $y=2(4)-3=5$ .

(4 marks)

Tier 3 · Hard

Solve

(x^2-5x)^2-5(x^2-5x)-14=0

, giving exact answers.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=\dfrac{5\pm\sqrt{53}}{2}$ $x=\dfrac{5\pm\sqrt{17}}{2}$	6
Notes
Let $u=x^2-5x$ . Then $u^2-5u-14=0$ , so $(u-7)(u+2)=0$ . If $u=7$ , then $x^2-5x-7=0$ , giving $x=(5\pm\sqrt{53})/2$ . If $u=-2$ , then $x^2-5x+2=0$ , giving $x=(5\pm\sqrt{17})/2$ . All four values satisfy the original equation.

(6 marks)

Tier 1 · Easy

State, with a reason, the number of real roots of

2x^2+3x+5=0

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	No real roots, because the discriminant is $-31<0$ .	2
Notes
The discriminant is $b^2-4ac=3^2-4(2)(5)=9-40=-31$ . Since it is negative, the equation has no real roots.

(2 marks)

Tier 2 · Standard

Find the set of real values of

k

for which

x^2+(k-3)x+k=0

has no real roots.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$1<k<9$	4
Notes
No real roots requires a negative discriminant. Thus $(k-3)^2-4k<0$ , which simplifies to $k^2-10k+9<0$ , or $(k-1)(k-9)<0$ . This upward-opening quadratic is negative between its roots, so $1<k<9$ .

(4 marks)

Tier 3 · Hard

Solve

25^x-26(5^x)+25=0

, giving exact values.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x=0$ or $x=2$	5
Notes
Let $u=5^x$ , where $u>0$ . Since $25^x=(5^x)^2$ , the equation becomes $u^2-26u+25=0=(u-1)(u-25)$ . Hence $5^x=1$ or $5^x=25$ , giving $x=0$ or $x=2$ .

(5 marks)

Tier 2 · Standard

A monic quadratic

f(x)=x^2+px+q

has roots

2+\sqrt3

and

2-\sqrt3

. Find

p

and

q

, and hence state the minimum value of

f(x)

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$p=-4$ , $q=1$ The minimum value is $-3$ .	4
Notes
The sum of the roots is $4$ , so $-p=4$ and $p=-4$ . Their product is $(2+\sqrt3)(2-\sqrt3)=1$ , so $q=1$ . Hence $f(x)=x^2-4x+1=(x-2)^2-3$ , whose minimum value is $-3$ .

(4 marks)

Tier 3 · Hard

Solve

\left(x+\dfrac1x\right)^2-5\left(x+\dfrac1x\right)+6=0

for real

x

, giving exact answers.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x=1$ or $x=\dfrac{3+\sqrt5}{2}$ or $x=\dfrac{3-\sqrt5}{2}$	6
Notes
The original expression requires $x\ne0$ . Let $u=x+1/x$ . Then $u^2-5u+6=0$ , so $u=2$ or $u=3$ . If $u=2$ , then $x^2-2x+1=0$ , giving $x=1$ . If $u=3$ , then $x^2-3x+1=0$ , giving $x=(3\pm\sqrt5)/2$ . All three values are non-zero and satisfy the original equation.

(6 marks)

Tier 3 · Hard

The roots of

2x^2-5x-3=0

are

\alpha

and

\beta

. Without solving this equation, find an equation with integer coefficients whose roots are

\alpha^2

and

\beta^2

. Give your equation in the form

ax^2+bx+c=0

, where

a>0

and

a

b

c

have greatest common divisor

1

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$4x^2-37x+9=0$	5
Notes
The original roots satisfy $\alpha+\beta=5/2$ and $\alpha\beta=-3/2$ . Hence $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=25/4+3=37/4$ , while $\alpha^2\beta^2=(\alpha\beta)^2=9/4$ . The required equation is $x^2-(37/4)x+9/4=0$ . Multiplying by $4$ gives the pinned integer form $4x^2-37x+9=0$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Find the exact range of

f(x)=\dfrac{x^2+4x+7}{x^2+1}

for real

x

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$4-\sqrt{13}\leq f(x)\leq4+\sqrt{13}$	6
Notes
Set $y=(x^2+4x+7)/(x^2+1)$ ; the denominator is positive for every real $x$ . Rearranging gives $(y-1)x^2-4x+(y-7)=0$ . For $y\ne1$ , this has a real solution exactly when its discriminant is non-negative: $16-4(y-1)(y-7)\geq0$ . This simplifies to $y^2-8y+3\leq0$ , so $4-\sqrt{13}\leq y\leq4+\sqrt{13}$ . The value $y=1$ is also attained, at $x=-3/2$ , and lies in this interval. At each endpoint the quadratic has a repeated real root, so both endpoints are included.

(6 marks)

Verified exam appearances

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Other points in 2 Algebra and functions

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