Revision for completing the square, maximum and minimum values, the discriminant, tangency and intersection conditions, quadratic inequalities, and (4PM1) the sum and product of the roots. Both boards. Only Edexcel 4PM1 asks for the sum and product of the roots (α and β) and forming new equations from them.
Completing the square: ax² + bx + c = a(x + b/2a)² + c − b²/4a. The vertex is at x = −b/2a.
The discriminant b² − 4ac decides the roots: positive, two distinct real roots; zero, two equal roots; negative, no real roots.
For a line and a curve, substitute to get one quadratic. Discriminant zero means the line is a tangent; negative means they do not meet.
For a quadratic inequality, find the critical values, sketch the parabola and read off the region. Answers are either between the roots or outside them.
For ax² + bx + c = 0 with roots α and β: α + β = −b/a and αβ = c/a. Write expressions such as α² + β² as (α + β)² − 2αβ.
Watch out: When a question says 'two distinct real roots' with a k in front of x², also say k ≠ 0, or the equation is no longer a quadratic.
Worked example· 4 marks
(a) Express \(2x^2 - 12x + 23\) in the form \(a(x + b)^2 + c\).
(b) Hence write down the coordinates of the minimum point of the curve \(y = 2x^2 - 12x + 23\).
B1 (a) a = 2
M1 (a) 2(x − 3)² − 18 + 23
A1 (a) 2(x − 3)² + 5
B1 (b) (3, 5)
Take out the 2 from the x terms: 2(x² − 6x) + 23.
Complete the square: x² − 6x = (x − 3)² − 9, so 2((x − 3)² − 9) + 23.
= 2(x − 3)² − 18 + 23 = 2(x − 3)² + 5.
The least value of 2(x − 3)² is 0, at x = 3, so the minimum point is (3, 5).
Answer: (a) \(2(x - 3)^2 + 5\) (b) \((3, 5)\)
Quadratics questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 3 marks06064PM1no calculator
Find the values of \(k\) for which the equation \(x^2 + kx + 16 = 0\) has two equal roots.
Show the mark scheme and worked solution
M1 b² − 4ac = 0 used
A1 k² − 64 = 0
A1 k = 8 or k = −8
Equal roots: the discriminant b² − 4ac is 0.
k² − 4 × 1 × 16 = 0, so k² = 64.
k = 8 or k = −8.
Answer: \(k = 8\) or \(k = -8\)
Question 2 · 4 marks06064PM1no calculator
The line \(y = 2x + c\) is a tangent to the curve \(y = x^2 - 4x + 7\).
Find the value of \(c\) and the coordinates of the point where the line touches the curve.
Show the mark scheme and worked solution
M1 x² − 6x + 7 − c = 0
M1 discriminant = 0: 36 − 4(7 − c) = 0
A1 c = −2
A1 (3, 4)
Set the equations equal: x² − 4x + 7 = 2x + c, so x² − 6x + 7 − c = 0.
A tangent meets the curve once: 36 − 4(7 − c) = 0.
36 − 28 + 4c = 0, so c = −2.
Then x² − 6x + 9 = 0, (x − 3)² = 0, x = 3 and y = 2(3) − 2 = 4.
Answer: \(c = -2\), the point \((3, 4)\)
Question 3 · 5 marks4PM1no calculator
The equation \(3x^2 - 4x - 2 = 0\) has roots \(\alpha\) and \(\beta\).
Without finding \(\alpha\) and \(\beta\), work out the exact value of
IGCSE Math Revision