IGCSE · Mathematics 0580 · Algebra & graphs

Quadratic equations: IGCSE Mathematics questions with answers

Free Quadratic equations questions for IGCSE Mathematics (0580), part of Algebra & graphs. Each question has the answer and the marking points an examiner would look for.

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Quadratic equations practice questions

Question 1Structured · 12 marksA* challenge

A rectangle has sides of length 1 and x, where x > 1. When a 1 by 1 square is removed from one end, the rectangle that remains is mathematically similar to the original rectangle.

(a)

Find the exact value of x.

[5]
(b)

Without using a calculator, find x3 in the form a + b√5.

[4]
(c)

Find x6 in the form a + b√5, and describe the pattern in the coefficients of xn = px + q.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x = 1 + √52
remaining rectangle is 1 by (x − 1) B1
x1 = 1x − 1 M1
x2 − x − 1 = 0 A1
formula M1
1 + √52 (rejecting the negative root) A1
5
(b)x³ = x·x² = x(x + 1) = x² + x = 2x + 1 = 2 + √5
x2 = x + 1 B1
x3 = x2 + x M1
= 2x + 1 A1
2 + √5 A1
4
(c)x⁶ = 8x + 5 = 9 + 4√5; p, q are consecutive Fibonacci numbers
repeated use of x2 = x + 1 (x4 = 3x + 2, x5 = 5x + 3) M1
x6 = 8x + 5 = 9 + 4√5 A1
coefficients are consecutive Fibonacci numbers B1
3
Question 2Structured · 6 marksA* challenge

A rectangle has length (x + 3) cm and width (2x − 3) cm. Its area is 52 cm².

(a)

Show that 2x2 + 3x − 61 = 0.

[2]
(b)

Solve the equation to find the perimeter of the rectangle. Give your answer correct to 2 decimal places.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)(x + 3)(2x − 3) = 52 → 2x2 + 3x − 9 = 52
correct expansion of brackets M1
correctly rearranged to given form with no errors A1
2
(b)x = 4.823…, perimeter = 28.94 cm
−3 ± √(32 − 4 × 2 × (−61))2 × 2 M1
x = 4.823… (reject negative) A1
perimeter = 2(x + 3) + 2(2x − 3) with their x M1
28.94 A1
4
Question 3Structured · 3 marks

Solve 4x2 − 27x + 18 = 0.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
x = 34 or x = 6
(4x − 3)(x − 6) = 0 B2
B1 for factors giving two of the three terms correctly
34 and 6 A1
3
Question 4Structured · 5 marksA* challenge
(a)

Write 3x2 − 24x + 53 in the form a(x + b)2 + c.

[3]
(b)

Write down the coordinates of the turning point of y = 3x2 − 24x + 53, and state whether it is a maximum or a minimum.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)3(x − 4)2 + 5
a = 3 B1
b = −4 B1
c = 5 B1
3
(b)(4, 5), minimum
(4, 5) B1
minimum B1
2
Question 5Structured · 10 marksA* challenge

The line y = mx − 6 is a tangent to the curve y = x2 + 3x − 5.

(a)

Find the two possible values of m.

[4]
(b)

For each value of m, find the coordinates of the point of contact.

[3]
(c)

Find the area of the triangle enclosed by the two tangents and the x-axis.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)m = 5 or m = 1
equate and rearrange: x2 + (3 − m)x + 1 = 0 M1
discriminant = 0 M1
(3 − m)2 = 4 A1
5, 1 A1
4
(b)(1, −1) and (−1, −7)
repeated root x = m − 32 M1
(1, −1) A1
(−1, −7) A1
3
(c)725
tangents meet at (0, −6) B1
x-intercepts 65 and 6 M1
725 A1
3
Question 6Structured · 6 marksA* challenge

A rectangle has length (x + 6) cm and width (2x − 5) cm. Its area is 84 cm².

(a)

Show that 2x2 + 7x − 114 = 0.

[2]
(b)

Solve the equation to find the perimeter of the rectangle.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)(x + 6)(2x − 5) = 84 → 2x2 + 7x − 30 = 84
correct expansion of brackets M1
correctly rearranged to given form with no errors A1
2
(b)x = 6, perimeter = 38 cm
correct factorisation or formula M1
x = 6 (reject negative root) A1
perimeter = 2(x + 6) + 2(2x − 5) with their x M1
38 A1
4

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in