Many real situations turn into quadratic equations: the height of a ball thrown at an angle, the size of a field with a given area, or two numbers with a given product.
A thrown ball flies along a parabola, so solving the quadratic equation with height set to 0 tells you when and where it lands.
The method is always the same: let x be the quantity you want, write an equation from the conditions, solve it, and finally check that the answer fits the conditions.
A quadratic equation often has two roots, and if the quantity cannot be negative, like a length or a count, the negative root must be rejected.
The parabola of a thrown ball
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The parabola of a ball, its greatest height and horizontal distance, as the initial speed and angle change
20 m/s
45°
🏀 Projectile motion and quadratic equations
①Written in terms of the horizontal distance x, the ball's height y is a quadratic of the form y = ax² + bx (a < 0).
②The ball lands where y = 0, so the landing point comes from solving ax² + bx = 0.
③The highest point is the vertex of the parabola.
④The picture uses g ≈ 10 m/s². At the starting values, 20 m/s and 45°, the horizontal distance is 40 m and the greatest height is 10 m.
Setting up an area problem
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The area of a 10 by 8 rectangle when both its width and its height are increased by x
3
📐 Steps for area problems
①Let x be the length added or removed.
②Write the area of the new shape in terms of x; in the picture it is (10 + x)(8 + x).
③Set up an equation from the area condition: an area of 120 gives x² + 18x − 40 = 0.
④This gives x = 2 or x = −20, but a length must be positive, so only x = 2 is the answer.
Problems about numbers
Consecutive natural numbers
n(n + 1) = 90 → n² + n − 90 = 0
Let the smaller of two consecutive natural numbers be n, so they are n and n + 1.
🔢 Common number problems
①If the product of two consecutive natural numbers is A, then n(n + 1) = A.
②If you are told the square of a number minus 3, write x² − 3 = (that value).
③A two-digit natural number with tens digit a and units digit b is written 10a + b.
Four steps for word problems
Order of work
choose the unknown → write the equation → solve it → check
Follow this order whatever the type of problem.
⚠️ Why the check matters
①A quadratic equation can have two roots.
②Any root that does not fit the conditions of the problem is rejected.
③In problems about natural numbers, lengths or times, negative roots are rejected.
④Put the answer back into the original situation to make sure it works.
Worked examples
Example 1
The product of two consecutive natural numbers is 56. Find them.
1
Let the smaller number be n; then n(n + 1) = 56.
n(n+1)=56 ⇒ n²+n−56=0
2
Factorising gives (n − 7)(n + 8) = 0, and n is a natural number, so n = 7.
(n−7)(n+8)=0 ⇒ n=7
▸ 7 and 8
The natural-number condition rejects the negative root n = −8.
Example 2
For a positive number x, x(x+3)=40. Find x.
1
Expand and rearrange so the right-hand side is 0.
x²+3x−40=0
2
(x − 5)(x + 8) = 0, and x > 0, so x = 5.
(x−5)(x+8)=0 ⇒ x=5
▸ x = 5
The positive condition rejects x = −8.
Exam checklist
Setting up by type
consecutive numbers: n, n + 1 · area: width × height · height of an object thrown up: h = v₀t − 5t²
Choosing the unknown well for each type is half the solution.
G9 school-exam type
An object is launched from the ground; its height after t seconds is h=−5t²+20t (m). After how many seconds does it return to the ground?
①1 s
②2 s
③4 s
④5 s
⑤8 s
▸ ③ 4 s
1
On the ground means h=0.
−5t²+20t=0
2
From −5t(t−4)=0 with t>0, t=4.
−5t(t−4)=0 ⇒ t=4
🎯 Exam points
①Let the quantity you want be the unknown.
②After writing the equation, rearrange so the right-hand side is 0.
③After finding the roots, check conditions such as natural or positive.
④Write the answer with its units.
⑤If the problem says "greater than" or "at most", check the condition again.