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Topic 2.07 · SL and HL

Solving roots of quadratics (factorising, CTS, quadratic formula) + discriminant: notes and practice questions

Summary
  • Factorising: Write ax2+bx+c=0ax^2 + bx + c = 0 as (px+q)(rx+s)=0(px + q)(rx + s) = 0.
  • Completing the square (CTS): Rewrite ax2+bx+cax^2 + bx + c as a(x−h)2+ka(x-h)^2 + k.
  • Quadratic formula: x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .
  • Discriminant (Δ=b2−4ac\Delta = b^2 - 4ac):
  • Δ>0\Delta > 0: Two distinct roots.
  • Δ=0\Delta = 0: One repeated root.
  • Δ<0\Delta < 0: No real roots.

How it is examined

The discriminant question with a parameter is the standard version, and it ends in an inequality that has to be solved and stated correctly. At SL "no real roots" means exactly that, not "two complex roots", because complex numbers are HL. `Find`, `Show that`, `Determine`. 5 to 7 marks.

Given in the booklet

Both the quadratic formula x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} and the discriminant Δ=b2−4ac\Delta = b^2 - 4ac are given.

Key ideas
  • Solution of quadratic equations and inequalities.
  • The quadratic formula.
  • The discriminant Δ=b2−4ac\Delta = b^2 - 4ac and the nature of the roots, that is, two distinct real roots, two equal real roots, no real roots.
At HL

Complex roots arrive at AHL 1.14; inequalities are generalised at AHL 2.15.

Linking questions

  • Links to other subjects: projectile motion and energy changes in simple harmonic motion (physics); equilibrium equations (chemistry).

Practice questions

94 questions · 2 easy · 69 medium · 23 hard
Showing 20 of 20

Question 1

EasyPaper 1 · no calculator5 marks
(a)

Consider the functions f(x)=x2−4f(x) = x^2 - 4 and g(x)=pxg(x) = \frac{p}{x}, where pp is a non-zero real constant.

(a) Write down an expression for (f∘g)(x)(f \circ g)(x).

[2]
(b)

(b) Given that (f∘g)(2)=5(f \circ g)(2) = 5, find the possible values of pp.

[3]

Question 2

MediumPaper 1 · no calculator5 marks

If the equation 2e2x+lnk=−4ex2e^{2x} + lnk = - 4e^{x} is said to have exactly one real solution, find the value of kk.

Question 3

HardPaper 1 · no calculator15 marks
(a)(i)

Consider the series sin⁡θ+ksin⁡θ+12sin⁡θ+… \sin\theta + k\sin\theta + \frac{1}{2}\sin\theta + \dots , where 0<θ<π2 0 < \theta < \frac{\pi}{2} and k∈R,k≠0 k \in \mathbb{R}, k \neq 0 .

Consider the case where the series is geometric.

(a) (i) Show that k=±12 k = \pm \frac{1}{\sqrt{2}} .

[2]
(a)(ii)

(a) (ii) Given that k>0 k > 0 and the sum to infinity is 2 \sqrt{2} , find the value of sin⁡θ \sin\theta .

[3]
(b)(i)

Now consider the case where the series is arithmetic with common difference dd.

(b) (i) Show that k=34 k = \frac{3}{4} .

[3]
(b)(ii)

(b) (ii) Write down dd in the form csin⁡θ c\sin\theta , where c∈Q c \in \mathbb{Q} .

[1]
(b)(iii)

(b) (iii) The sum of the first nn terms of the series is −14sin⁡θ -14\sin\theta .

Find the value of n n .

[6]

Question 4

EasyPaper 2 · calculator4 marks

Consider a function ff which has a first derivative given by f′(x)=5−2x−x2f'(x) = 5 - 2x - x^2, where x∈Rx \in \mathbb{R}.

Find the range of values of xx for which ff is increasing.

Question 5

MediumPaper 1 · no calculator8 marks
(a)

Consider the function g(x)=ax2+x+kx−4g(x) = \frac{ax^2+x+k}{x-4}.

The graph of y=g(x) y=g(x) passes through the point (1,−2) (1, -2) and has an oblique asymptote with equation y=−3x−11 y = -3x-11 .

(a) Write down the equation of the vertical asymptote.

[1]
(b)(i)

(b) Find the value of:

(i) aa

[2]
(b)(ii)

(ii) kk

[2]
(c)

(c) Hence, find the exact coordinates of any points where the graph of y=g(x)y=g(x) intersects the x-axis.

[3]

Question 6

HardPaper 1 · no calculator19 marks
(a)

Let f(x)=11−2xf(x) = \frac{1}{\sqrt{1-2x}} for x<12x < \frac{1}{2}.

(a) Show that f′′(x)=3(1−2x)−52f''(x) = 3(1-2x)^{-\frac{5}{2}}.

[3]
(b)

(b) Use mathematical induction to prove that f(n)(x)=(2n)!2nn!(1−2x)−2n+12f^{(n)}(x) = \frac{(2n)!}{2^n n!} (1-2x)^{-\frac{2n+1}{2}} for n∈Z,n≥2n \in \mathbb{Z}, n \ge 2.

[9]
(c)

Let g(x)=ln⁡(1+kx)g(x) = \ln(1+kx), where kk is a real constant.

Consider the function hh defined by h(x)=f(x)×g(x)h(x) = f(x) \times g(x) for x<12x < \frac{1}{2}.

It is given that the coefficient of the x2x^2 term in the Maclaurin series for h(x)h(x) is −4-4.

(c) Find the possible values of kk.

[7]

Question 7

MediumPaper 2 · calculator6 marks

Events S and C are independent. The probability that a student passes a Statistics exam, P(S), is twice the probability that the student passes a Calculus exam, P(C).

Given that the probability a student passes at least one of these exams is 0.625, find the probability that the student passes the Calculus exam, P(C).

Question 8

HardPaper 1 · no calculator19 marks
(a)

Two spacecraft, S1 and S2, travel along straight paths, represented by the lines L1L_1 and L2L_2 respectively. The paths of the spacecraft intersect at a docking station D. A probe is located at a point P on the path of L2L_2. This is shown in the following diagram.

Diagram showing two intersecting lines L1 and L2, with point D at the intersection and point P on L2

The direction vector of L1L_1 is (21−2)\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}. The vector DP⃗\vec{\text{DP}} is given by (k40)\begin{pmatrix} k \\ 4 \\ 0 \end{pmatrix}, where k≥0k \ge 0.

The acute angle between the paths L1L_1 and L2L_2 is θ\theta, where cos⁡θ=13\cos\theta = \frac{1}{3}.

(a) Show that 2k+4=k2+162k+4 = \sqrt{k^2+16}.

[4]
(b)

(b) Find the value of kk.

[3]
(c)

(c) Hence, find the shortest distance from the probe at P to the path L1L_1.

[3]
(d)

The paths L1L_1 and L2L_2 lie on a plane, Π\Pi.

(d) Find a vector normal to the plane Π\Pi.

[2]
(e)

A satellite dish is modelled as a right circular cone with its vertex at V. The base of the cone lies in the plane Π\Pi and is centred at P. The path L1L_1 is tangent to the circular base of the cone. The volume of the cone is 128π29\frac{128\pi\sqrt{2}}{9} cubic units. The position vector of P is (123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}.

(e) Find the two possible position vectors for V.

[7]

Question 9

MediumPaper 1 · no calculator7 marks
(a)

(a) Show that 2x+5+3x−1=2x2+3x−2x−12x+5 + \frac{3}{x-1} = \frac{2x^2 + 3x - 2}{x-1}, for x∈R,x≠1x \in \mathbb{R}, x \neq 1.

[2]
(b)

(b) Hence or otherwise, solve the equation 2sin⁡θ+5+3sin⁡θ−1=02\sin{\theta} + 5 + \frac{3}{\sin{\theta}-1} = 0 for 0≤θ≤2π0 \leq \theta \leq 2\pi, θ≠π2\theta \neq \frac{\pi}{2}.

[5]

Question 10

HardPaper 1 · no calculator6 marks

Find the set of values for the constant kk such that the equation 2cos⁡2θ+5cos⁡θ=k−12\cos^2\theta + 5\cos\theta = k-1 has at least one real solution for θ\theta.

Question 11

MediumPaper 1 · no calculator7 marks
(a)

The graphs of the functions f(x)=kx2−3xf(x) = kx^2 - 3x and g(x)=x−kg(x) = x - k intersect at two distinct points.

(a) Find the set of possible values for kk.

[5]
(b)

(b) Consider the case when k=1k=1. The x-coordinates of the intersection points can be written in the form x=m±nx = m \pm \sqrt{n}, where m,n∈Zm, n \in \mathbb{Z}. Find the values of mm and nn.

[2]

Question 12

HardPaper 1 · no calculator14 marks
(a)

A rectangle is inscribed in an ellipse with equation x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1. The sides of the rectangle are parallel to the coordinate axes. The vertices of the rectangle are located at (±x,±y)(\pm x, \pm y), where x>0x > 0 and y>0y > 0.

Diagram of an ellipse with an inscribed rectangle

(a) Show that the area of the rectangle, AA, can be expressed as A=12x525−x2A = \frac{12x}{5}\sqrt{25-x^2}.

[4]
(b)

(b) Show that dAdx=12(25−2x2)525−x2\frac{dA}{dx} = \frac{12(25-2x^2)}{5\sqrt{25-x^2}}.

[4]
(c)

(c) Hence, find the exact dimensions of the rectangle with the maximum possible area.

[6]

Question 13

MediumPaper 1 · no calculator7 marks

A solid is formed by rotating the curve with equation x2=8yx^2 = 8y for y≥0y \ge 0 by 360∘360^\circ about the yy-axis. This forms a solid paraboloid.

A cylindrical hole of radius 44 is drilled through the center of the paraboloid, along the yy-axis. The resulting solid is a ring of height hh.

This information is shown in the following diagrams.

Diagram showing a cross-section of a paraboloid with a cylindrical hole and a 3D view of the resulting ring

The volume of the ring is 144π144\pi.

Find the value of hh.

Question 14

HardPaper 1 · no calculator17 marks
(a)

Find the binomial expansion of (cos⁡θ+isin⁡θ)4(\cos \theta + i \sin \theta)^4. Give your answer in the form a+bia + bi where aa and bb are expressed in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

[4]
(b)

By using De Moivre's theorem and your answer to part (a), show that cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1.

[6]
(c)(i)

Hence, show that θ=π8\theta = \frac{\pi}{8} and θ=3π8\theta = \frac{3\pi}{8} are solutions of the equation 8cos⁡4θ−8cos⁡2θ+1=08\cos^4\theta - 8\cos^2\theta + 1 = 0.

[3]
(c)(ii)

Hence, find the exact value of cos⁡(π8)cos⁡(3π8)\cos(\frac{\pi}{8})\cos(\frac{3\pi}{8}).

[4]

Question 15

MediumPaper 1 · no calculator13 marks
(a)

The functions ff and gg are defined by

f(x)=2ln⁡xf(x) = 2\ln x, where x>0x > 0

g(x)=ln⁡(k(x−2))g(x) = \ln(k(x-2) ), where x>2,k∈R+x > 2, k \in \mathbb{R}^+.

The graphs of y=f(x)y = f(x) and y=g(x)y = g(x) intersect at two distinct points.

(a) State the equation of the vertical asymptote to the graph of y=g(x)y = g(x).

[1]
(b)(i)

(b) (i) Show that, at the points of intersection, x2−kx+2k=0x^2 - kx + 2k = 0.

[3]
(b)(ii)

(b) (ii) Hence show that k2−8k>0k^2 - 8k > 0.

[2]
(b)(iii)

(b) (iii) Find the range of possible values of kk.

[2]
(c)

The graphs intersect at x=px=p and x=qx=q, where p<qp<q.

(c) In the case where k=10k=10, find the value of q−pq-p. Express your answer in the form aba\sqrt{b}, where a,b∈Z+a, b \in \mathbb{Z}^+.

[5]

Question 16

HardPaper 1 · no calculator15 marks
(a)

A drone takes off from a platform. Its height, hh metres, above the platform after tt seconds is given by h(t)=6t−t2h(t) = 6t - t^2, for 0≤t≤80 \le t \le 8. This is shown in the following diagram.

Graph of height h versus time t for the drone, showing a parabola opening downwards with vertex in the first quadrant and passing through the origin

The drone lands back on the platform when t=pt=p.

Find the value of pp.

[2]
(b)(i)

The drone reaches its maximum height when t=qt=q.

Find the value of qq.

[3]
(b)(ii)

Find the drone's maximum height above the platform.

[2]
(c)

Find the drone's vertical distance from the platform when t=8t=8.

[2]
(d)

The total vertical distance travelled by the drone in the first 8 seconds is given by dd.

Find the value of dd.

[2]
(e)

A second drone, Drone B, takes off from the same platform. Its velocity is given by vB(t)=8−2tv_B(t) = 8 - 2t, for t≥0t \ge 0.

When t=kt = k, the total vertical distance travelled by Drone B is equal to dd.

Find the value of kk.

[4]

Question 17

MediumPaper 1 · no calculator15 marks
(a)

Consider the function ff defined by f(x)=x3−6x2+8xf(x) = x^3 - 6x^2 + 8x.

Find the xx-intercepts of the graph of y=f(x)y=f(x).

[3]
(b)

The graph of y=f(x)y=f(x) for 0≤x≤40 \le x \le 4 is shown below. The graph encloses two regions with the xx-axis, shaded in the diagram.

Graph of y = x^3 - 6x^2 + 8x from x=0 to x=4, showing two regions bounded by the x-axis. The first region from x=0 to x=2 is above the axis, the second from x=2 to x=4 is below the axis.

Find the total area of the shaded regions.

[6]
(c)

The total surface area of a closed right cylinder is 8, equal to the total shaded area found in part (b). The cylinder has a height of 4−ππ\frac{4-\pi}{\pi}.

Diagram of a cylinder with radius r and height h.

Find the radius, rr, of the cylinder.

[4]
(d)

Hence, find the volume of the cylinder.

[2]

Question 18

HardPaper 1 · no calculator9 marks
(a)

A function ff is defined by f(x)=4x−12x+3f(x) = \frac{4x-1}{2x+3}, where x∈R,x≠−32x \in \mathbb{R}, x \neq -\frac{3}{2}.

The graph of y=f(x)y = f(x) is shown below.

Graph of the function f(x) showing its two branches and asymptotes.

(a) Write down the equation of the horizontal asymptote.

[1]
(b)(i)

Consider the function g(x)=mx−13g(x) = mx - \frac{1}{3}, where m∈R,m≠0m \in \mathbb{R}, m \neq 0.

(i) Write down the number of solutions to f(x)=g(x)f(x) = g(x) for m<0m < 0.

[1]
(b)(ii)

(ii) Determine the value of mm such that f(x)=g(x)f(x) = g(x) has only one solution for xx.

[4]
(b)(iii)

(iii) Determine the range of values for mm for which f(x)=g(x)f(x) = g(x) has two distinct solutions for x≤0x \le 0.

[3]

Question 19

MediumPaper 1 · no calculator8 marks
(a)

The functions ff and gg are defined by f(x)=sin⁡xf(x) = \sin x and g(x)=cot⁡xg(x) = \cot x, for 0<x<π20 < x < \frac{\pi}{2}.

The curves y=f(x)y = f(x) and y=g(x)y = g(x) intersect at a point P whose x-coordinate is kk.

Show that sin⁡2k=cos⁡k\sin^2 k = \cos k.

[2]
(b)

Hence, show that the tangent to the curve y=f(x)y = f(x) at P and the tangent to the curve y=g(x)y = g(x) at P are perpendicular.

[3]
(c)

Find the value of cos⁡k\cos k. Give your answer in the form a+bc\frac{a+\sqrt{b}}{c}, where a,c∈Za, c \in \mathbb{Z} and b∈Z+b \in \mathbb{Z}^+.

[3]

Question 20

HardPaper 1 · no calculator7 marks
(a)

A project manager has nn employees available for a new project. The employees are to be divided into two teams, Team Alpha and Team Beta. For the project to be successful, Team Alpha must have exactly four members and Team Beta must have at least four members.

The manager will randomly assign four employees to Team Alpha, with the rest forming Team Beta.

Write down an expression for the number of ways that the teams could be formed.

[1]
(b)

Two of the employees, Chloe and David, have a history of conflict and cannot be in the same team. The manager agrees to this condition, and finds that this restriction reduces the number of possible team formations to two-fifths of the original number.

Determine the value of nn.

[6]

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What does Solving roots of quadratics (factorising, CTS, quadratic formula) + discriminant cover in IB Maths AA?

Factorising: Write ax^2 + bx + c = 0 as (px + q)(rx + s) = 0. Completing the square (CTS): Rewrite ax^2 + bx + c as a(x-h)^2 + k. Quadratic formula: x = frac-b ± √b^2 - 4ac2a.

Is Solving roots of quadratics (factorising, CTS, quadratic formula) + discriminant SL or HL?

Both. SL and HL students study Solving roots of quadratics (factorising, CTS, quadratic formula) + discriminant, and HL goes further: Complex roots arrive at AHL 1.14; inequalities are generalised at AHL 2.15.

How do I revise Solving roots of quadratics (factorising, CTS, quadratic formula) + discriminant for IB Maths AA?

Start from the core idea: factorising: Write ax^2 + bx + c = 0 as (px + q)(rx + s) = 0. In the exam: the discriminant question with a parameter is the standard version, and it ends in an inequality that has to be solved and stated correctly. At SL "no real roots" means exactly that, not "two complex roots", because complex numbers are HL. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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