IB Class 9 Quadratic Equations Practice Questions

IB Class 9 Quadratic Equations Practice Questions

IB Class 9 Quadratic Equations Practice Questions help students understand factorisation and quadratic formula methods clearly. The exercises focus on solving equations step by step. Moreover, learners practice completing the square and analyzing roots. This strengthens algebraic reasoning and accuracy.

Build Strong Algebra Foundations

IB Class 9 Quadratic Equations Practice Questions support effective exam revision. Therefore, students gain confidence while solving word problems and equation systems. Additionally, consistent practice improves speed and precision. As a result, learners perform better in assessments and strengthen long-term mathematical skills.

Quadratic Worksheets and Notes

Students can use IB Math Quadratic Equations Class 9 Notes PDF for concept clarity. Moreover, Quadratic Equations IB Grade 9 Worksheet PDF provides structured exercises. Therefore, regular revision improves understanding and boosts exam performance effectively.

Quadratic Equations (Introduction)

Quadratic equations are polynomial equations of degree 2. They appear in many real-world contexts, from physics (projectile motion) to economics (profit maximization) and geometry (area problems). This chapter introduces their standard form, methods of solution, and the powerful quadratic formula.

1. What is a Quadratic Equation?

A quadratic equation is an equation that can be written in the standard form:

ax² + bx + c = 0    where a, b, c are constants and a ≠ 0
  • a is the coefficient of x² (quadratic term)
  • b is the coefficient of x (linear term)
  • c is the constant term

The condition a ≠ 0 is essential; if a = 0, the equation becomes linear (bx + c = 0).

Examples of Quadratic Equations

Example 1: 2x² + 5x – 3 = 0   (a = 2, b = 5, c = -3)

Example 2: x² – 7x = 0   (a = 1, b = -7, c = 0)

Example 3: 4x² – 9 = 0   (a = 4, b = 0, c = -9)   (This is called a pure quadratic)

Example 4: (x – 3)(x + 2) = 0   (This is a quadratic in factored form)

Note: The highest power of x is 2, which is why it’s called “quadratic” (from Latin “quadratus” meaning square).

2. Solving by Factorization

If a quadratic expression can be factored easily, this is often the quickest method. We use the zero product property: If the product of two factors is zero, then at least one of the factors must be zero.

Step-by-Step Process

  1. Write the equation in standard form (ax² + bx + c = 0).
  2. Factor the quadratic expression into two linear factors.
  3. Set each factor equal to zero and solve for x.

Example 1: Solve x² + 5x + 6 = 0

x² + 5x + 6 = (x + 2)(x + 3) = 0
x + 2 = 0   →   x = -2
x + 3 = 0   →   x = -3

Solutions: x = -2 or x = -3

Example 2: Solve 2x² – 7x + 3 = 0

2x² – 7x + 3 = (2x – 1)(x – 3) = 0
2x – 1 = 0   →   x = ½
x – 3 = 0   →   x = 3

Solutions: x = ½ or x = 3

Example 3: Solve x² – 9 = 0 (difference of squares)

x² – 9 = (x – 3)(x + 3) = 0
x – 3 = 0 → x = 3   or   x + 3 = 0 → x = -3

Solutions: x = 3 or x = -3

Remember: Not all quadratics factor nicely with integers. In such cases, we use other methods.

3. Solving Pure Quadratics (Square Root Method)

When a quadratic has no x-term (b = 0), it is called a pure quadratic. We can solve it by isolating x² and taking square roots.

Method

ax² + c = 0 → x² = -c/a → x = ±√(-c/a)

Remember to include both the positive and negative square roots.

Example 1: Solve 2x² – 32 = 0

2x² = 32
x² = 16
x = ±√16 = ±4

Solutions: x = 4 or x = -4

Example 2: Solve 3x² + 12 = 0

3x² = -12
x² = -4
x = ±√(-4) = ±2i   (where i = √-1)

Solutions: x = 2i or x = -2i (complex roots)

At Grade 9 level, we usually focus on real solutions. If the number under the square root is negative, we say “no real solutions”.

Example 3: Solve (x – 3)² = 25

x – 3 = ±√25 = ±5
x – 3 = 5 → x = 8
x – 3 = -5 → x = -2

Solutions: x = 8 or x = -2

4. Completing the Square

This method transforms any quadratic into a perfect square trinomial plus a constant. It is the foundation for deriving the quadratic formula.

Step-by-Step Process

For an equation ax² + bx + c = 0 (with a = 1 for simplicity):

  1. Move the constant term to the right side: x² + bx = -c
  2. Add (b/2)² to both sides (completing the square)
  3. Factor the left side as (x + b/2)²
  4. Solve using square roots

Example 1: Solve x² + 6x + 5 = 0 by completing the square.

x² + 6x = -5
Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4
(x + 3)² = 4
x + 3 = ±2
x = -3 + 2 = -1   or   x = -3 – 2 = -5

Solutions: x = -1 or x = -5

Example 2: Solve x² – 8x + 3 = 0.

x² – 8x = -3
Add (-8/2)² = 16: x² – 8x + 16 = 13
(x – 4)² = 13
x – 4 = ±√13
x = 4 ± √13

Solutions: x = 4 + √13 or x = 4 – √13

If a ≠ 1, first divide the entire equation by a before completing the square.

5. The Quadratic Formula

The quadratic formula is a universal method that works for all quadratic equations. It is derived by completing the square on the general form ax² + bx + c = 0.

x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}

Using the Formula

  1. Identify a, b, and c from the equation in standard form.
  2. Substitute into the formula.
  3. Simplify under the square root (the discriminant) first.
  4. Calculate the two values.

Example 1: Solve 2x² – 5x + 3 = 0 using the quadratic formula.

a = 2, b = -5, c = 3
x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4(2)(3)}}{2(2)}
x = \frac{5 \pm \sqrt{25 – 24}}{4} = \frac{5 \pm \sqrt{1}}{4}
x = \frac{5 + 1}{4} = \frac{6}{4} = 1.5   or   \frac{5 – 1}{4} = \frac{4}{4} = 1

Solutions: x = 1.5 or x = 1

Example 2: Solve x² + 4x + 2 = 0.

a = 1, b = 4, c = 2
x = \frac{-4 \pm \sqrt{16 – 8}}{2} = \frac{-4 \pm \sqrt{8}}{2}
x = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2}

Solutions: x = -2 + √2 or x = -2 – √2

6. The Discriminant (Δ or D)

The expression under the square root in the quadratic formula, b² – 4ac, is called the discriminant. It tells us about the nature of the roots without solving completely.

Nature of Roots

  • If b² – 4ac > 0: Two distinct real roots.
  • If b² – 4ac = 0: One real root (repeated root) – the parabola touches the x-axis.
  • If b² – 4ac < 0: No real roots (two complex roots).

Example 1: Determine the nature of roots for 2x² – 3x + 5 = 0.

Δ = (-3)² – 4(2)(5) = 9 – 40 = -31 < 0

∴ No real roots (complex roots).

Example 2: Find k such that x² – 6x + k = 0 has a repeated root.

For repeated root, Δ = 0
(-6)² – 4(1)(k) = 36 – 4k = 0
4k = 36 → k = 9

Answer: k = 9

7. Applications of Quadratic Equations

Solved Word Problems

Example 1 (Area): A rectangular garden has length 4 meters more than its width. If its area is 60 m², find the dimensions.

Let width = w, then length = w + 4
Area = w(w + 4) = 60
w² + 4w – 60 = 0
(w + 10)(w – 6) = 0
w = -10 (not possible) or w = 6

Dimensions: Width = 6 m, Length = 10 m

Example 2 (Projectile Motion): A ball is thrown upward with an initial velocity of 20 m/s. Its height h (in meters) after t seconds is given by h = 20t – 5t². When does the ball hit the ground?

Ground means h = 0: 20t – 5t² = 0
5t(4 – t) = 0
t = 0 (start) or t = 4

Answer: The ball hits the ground after 4 seconds.

Example 3 (Number Problem): Find two consecutive positive integers whose product is 132.

Let integers be n and n+1
n(n+1) = 132
n² + n – 132 = 0
(n + 12)(n – 11) = 0
n = -12 (reject) or n = 11

Integers: 11 and 12

8. Common Pitfalls

  • Incorrect: Forgetting to set the equation to zero before factoring.
    ✓ Correct: Always write in the form ax² + bx + c = 0 first.
  • Incorrect: Losing the ± sign when taking square roots.
    ✓ Correct: x² = 9 → x = ±3, not just x = 3.
  • Incorrect: Misidentifying a, b, c in the quadratic formula (especially signs).
    ✓ Correct: For 2x² – 3x – 5 = 0, a = 2, b = -3, c = -5.
  • Incorrect: Forgetting that if a ≠ 1, divide before completing the square.
    ✓ Correct: 2x² + 8x + 3 = 0 → divide by 2 first: x² + 4x + 1.5 = 0.
  • Incorrect: Thinking the discriminant determines the value, not just the nature.
    ✓ Correct: Discriminant only tells us about the type of roots (real/equal/complex).

9. Practice Questions

  1. Solve by factoring: x² – 7x + 12 = 0.
  2. Solve by square roots: 3x² – 27 = 0.
  3. Solve by completing the square: x² + 8x – 5 = 0.
  4. Use the quadratic formula: 2x² – 5x – 3 = 0.
  5. Find the discriminant and nature of roots for 4x² – 4x + 1 = 0.
  6. The product of two consecutive odd numbers is 143. Find the numbers.

Answers: 1) x = 3, 4 2) x = ±3 3) x = -4 ± √21 4) x = 3, -0.5 5) Δ = 0, one real root 6) 11 and 13

IB Mathematics – Grade 9

Quadratic Equations (Level 1)

Instructions:
  • Focus on translating the word problem into a quadratic expression.
  • Each question is worth 1 mark.
  • Detailed logic is available in the solutions section below.
1. A garden’s length is 5m more than its width (x). Area = 84 m². The equation is:
A) x² + 5x – 84 = 0
B) x² – 5x – 84 = 0
C) x² + 5x + 84 = 0
D) x² – 5x + 84 = 0
Answer: A
2. Product of two consecutive odd integers is 143. If x is the smaller:
A) x(x + 2) = 143
B) x(x – 2) = 143
C) x² – 2x = 143
D) x² + 2x = 143
Answer: A
3. A square has side x. Area is 144 m².
A) x² = 144
B) x² = 12
C) x² + 144 = 0
D) x² – 144 = 0
Answer: A
4. Ball height h = -5t² + 12t + 2. Equation for hitting the ground (h=0):
A) -5t² + 12t + 2 = 0
B) -5t² + 12t – 2 = 0
C) 5t² – 12t – 2 = 0
D) 5t² + 12t + 2 = 0
Answer: A
5. Sum of squares of two consecutive integers is 61. x is the smaller:
A) x² + (x + 1)² = 61
B) x² + (x – 1)² = 61
C) x² + x² = 61
D) x² + (x + 2)² = 61
Answer: A
6. Triangular flag base is 4m longer than height (x). Area = 20 m².
A) ½x(x + 4) = 20
B) ½x(x – 4) = 20
C) x(x + 4) = 20
D) x(x – 4) = 20
Answer: A
7. Pool is 10m long, x meters wide. Area = 120 m².
A) 10x = 120
B) x² = 120
C) 10x² = 120
D) x² + 10x – 120 = 0
Answer: A
8. Difference between a number and its square is 30.
A) x – x² = 30
B) x² – x = 30
C) x² + x = 30
D) x² – x + 30 = 0
Answer: B
9. Rocket height h = -5t² + 40t. Hits ground when:
A) -5t² + 40t = 0
B) -5t² – 40t = 0
C) 5t² + 40t = 0
D) 5t² – 40t = 0
Answer: A
10. Perimeter = 80m. Length is 10m more than width (x). Linear setup:
A) 2x(x + 10) = 80
B) x(x + 10) = 80
C) 2x + 2(x + 10) = 80
D) x² + 10x – 40 = 0
Answer: C
11. Product of two consecutive even integers is 168.
A) x(x + 2) = 168
B) x(x – 2) = 168
C) x² + 2x = 168
D) x² – 2x = 168
Answer: A
12. Square garden area = 225 m². Equation for side x:
A) x² = 225
B) x² = 15
C) x² + 225 = 0
D) x² – 225 = 0
Answer: A
13. Ball dropped from 45m: h = -5t² + 45. Hits ground when:
A) -5t² + 45 = 0
B) -5t² – 45 = 0
C) 5t² + 45 = 0
D) 5t² – 45 = 0
Answer: A
14. Land length is twice the width (x). Area = 32 m².
A) x(2x) = 32
B) 2x² = 32
C) x² = 32
D) x² + 2x – 32 = 0
Answer: A
15. Sum of squares of two consecutive odd integers is 130.
A) x² + (x + 2)² = 130
B) x² + (x – 2)² = 130
C) x² + x² = 130
D) x² + (x + 4)² = 130
Answer: A

Detailed Step-by-Step Solutions

1: Area = L × W. x(x+5) = 84 → x² + 5x – 84 = 0.
2: Consecutive odd numbers differ by 2. x and x+2. Product = x(x+2).
3: Square area is side squared. x² = 144.
4: Ground level is height zero. Set the provided equation h = 0.
5: Consecutive integers are x and x+1. Squares are x² and (x+1)².
6: Triangle Area = ½ × base × height. ½(x+4)(x) = 20.
7: Area = length × width. 10 × x = 120. (Linear in this case).
8: “Difference between square (x²) and number (x)” is x² – x = 30.
9: Impact time is when height h = 0. -5t² + 40t = 0.
10: Perimeter = 2(L + W). 2(x + x + 10) = 80.
11: Consecutive even integers differ by 2. x(x+2) = 168.
12: Area = x². x² = 225.
13: Set h = 0. -5t² + 45 = 0.
14: Width = x, Length = 2x. Area = x(2x) = 32.
15: Consecutive odd integers are x and x+2. Sum of squares = x² + (x+2)².

IB Mathematics – Difficult Level

Quadratic Equations Assessment

Teacher’s Notes:
  • Focus on multi-step modeling (Area with borders, 3D volume).
  • Questions involve both standard form $ax^2 + bx + c = 0$ and factored form.
  • Use the detailed solutions to check your algebraic expansions.
1. A pool is 12m long and $x$m wide. A uniform 1m border is added around it. Original Area = 108m², New Area = 160m².
  • A) x² + 12x – 108 = 0
  • B) x² + 12x – 160 = 0
  • C) (x + 2)(x + 12) = 160
  • D) (x + 2)(x + 12) = 108
Correct Answer: C
2. The product of two consecutive even integers is 288. Let $x$ be the smaller integer.
  • A) x(x + 2) = 288
  • B) x(x – 2) = 288
  • C) x² + 2x – 288 = 0
  • D) x² – 2x – 288 = 0
Correct Answer: A
3. A ball is thrown up at 24 m/s. Height $h = -5t^2 + 24t$. Equation for hitting the ground:
  • A) -5t² + 24t = 0
  • B) -5t² – 24t = 0
  • C) 5t² + 24t = 0
  • D) 5t² – 24t = 0
Correct Answer: A
4. A right-angled triangle has legs $x$ and $(x + 7)$. Hypotenuse is 13m.
  • A) x² + (x + 7)² = 13²
  • B) x² + (x – 7)² = 13²
  • C) x² + (x + 7) = 13²
  • D) x² + (x + 13)² = 7²
Correct Answer: A
5. Perimeter = 100m. Length is 10m more than twice the width ($x$). Find the width equation:
  • A) 2x(2x + 10) = 100
  • B) 2x + 2(2x + 10) = 100
  • C) x(2x + 10) = 100
  • D) x² + 10x – 25 = 0
Correct Answer: B
6. Sum of the squares of two consecutive odd integers is 290.
  • A) x² + (x + 2)² = 290
  • B) x² + (x – 2)² = 290
  • C) x² + (x + 4)² = 290
  • D) x² + (x + 1)² = 290
Correct Answer: A
7. A square piece of metal (side $x$). A 2cm strip is cut from ALL sides. Remaining area = 25cm².
  • A) (x – 4)² = 25
  • B) (x – 2)² = 25
  • C) x² – 4x + 4 = 25
  • D) x² – 2x + 4 = 25
Correct Answer: A
8. Rocket height $h = -5t^2 + 50t$. Find time when it hits ground ($h=0$).
  • A) -5t² + 50t = 0
  • B) -5t² – 50t = 0
  • C) 5t² + 50t = 0
  • D) 5t² – 50t = 0
Correct Answer: A
9. Box: Length = width($x$) + 5; Height = width($x$) – 2. Volume = 192cm³.
  • A) x(x + 5)(x – 2) = 192
  • B) x(x – 5)(x + 2) = 192
  • C) x(x + 5)(x + 2) = 192
  • D) x(x – 5)(x – 2) = 192
Correct Answer: A
10. The difference between the squares of two consecutive integers is 45.
  • A) (x + 1)² – x² = 45
  • B) x² – (x + 1)² = 45
  • C) (x + 1)² + x² = 45
  • D) x² + (x – 1)² = 45
Correct Answer: A
11. Flower bed (width $x$, length $x+3$) surrounded by a 1m path. Total area = 72m².
  • A) (x + 3 + 2)(x + 2) = 72
  • B) (x + 2)(x + 4) = 72
  • C) (x + 3)(x + 1) = 72
  • D) (x + 1)(x + 3 + 2) = 72
Correct Answer: A
12. Ball dropped from 80m. Height $h = -5t^2 + 80$. hits ground when:
  • A) -5t² + 80 = 0
  • B) -5t² – 80 = 0
  • C) 5t² + 80 = 0
  • D) 5t² – 80 = 0
Correct Answer: A
13. Cardboard box problem: Width $x$, Length $x+5$. 2cm squares cut from corners. Volume = 120cm³.
  • A) (x – 4)(x + 1)(2) = 120
  • B) (x – 4)(x + 5)(2) = 120
  • C) (x – 4)(x – 1)(2) = 120
  • D) (x – 2)(x + 3)(2) = 120
Correct Answer: A
14. The sum of a number and its square is 90.
  • A) x + x² = 90
  • B) x² + x = 90
  • C) x² – x = 90
  • D) x² + x – 90 = 0
Correct Answer: A
15. Triangular sail: Base = 2 $\times$ Height($x$). Area = 54m².
  • A) ½x(2x) = 54
  • B) ½x(x) = 54
  • C) x(2x) = 54
  • D) x² = 54
Correct Answer: A

Logic & Breakdown

1. New Length = 12+1+1=14. New Width = x+1+1=x+2. Area = 14(x+2)=160. This matches (x+12+2) logic.
4. Pythagorean Theorem: $a^2 + b^2 = c^2$. Hence $x^2 + (x+7)^2 = 13^2$.
7. If you cut 2cm from all sides, you lose 2cm on the left AND 2cm on the right. Side = x – 4. Area = (x-4)².
11. Border adds 2m to both length and width. Length: (x+3)+2. Width: x+2.
13. Folding up 2cm means New Width = x-4, New Length = (x+5)-4 = x+1, Height = 2. Volume = L $\times$ W $\times$ H.

Frequently Asked Questions (FAQs)

What are IB Class 9 Quadratic Equations Practice Questions?

They provide structured exercises on solving quadratic equations using different algebraic methods.

How do IB Class 9 Quadratic Equations Practice Questions help students?

IB Class 9 Quadratic Equations Practice Questions improve conceptual clarity and strengthen algebra skills.

Are quadratic formula methods included?

Yes, factorisation, completing the square, and formula methods are covered in IB Class 9 Quadratic Equations Practice Questions.

Do these questions in IB Class 9 Quadratic Equations Practice Questions follow IB curriculum standards?

Yes, they align with IB Grade 9 guidelines.

Are word problems included in the practice set?

Yes, real-life applications are included.

Can students use these questions for exam preparation?

Yes, they support systematic revision.

Are detailed solutions provided?

Yes, step-by-step explanations are included.

Are worksheets available in PDF format?

Yes, Quadratic Equations IB Grade 9 Worksheet PDF is available.

Do these resources include revision notes?

Yes, IB Math Quadratic Equations Class 9 Notes PDF can be used for revision.

How often should students practice quadratic equations?

Regular practice improves confidence and accuracy.