Quadratic Equations
Why This Matters
Suppose a charity trust wants a prayer hall with a carpet area of 300 square metres, and the length must be one metre more than twice the breadth. What dimensions work? If the breadth is x metres, the length is (2x + 1) metres, so the area is x(2x + 1) = 2x² + x = 300. Notice the x² — this isn’t a straight-line equation any more. You can’t just rearrange and divide your way out; the squared term changes everything.
Equations like this — where the highest power of the unknown is 2 — are quadratic equations, and they’re everywhere: the path of a thrown ball, the area of a field, the speed of a train that would arrive earlier if it sped up, profit that depends on price. In Chapter 3 each equation drew a straight line. Here, one equation drives a parabola (a U-shaped curve), and that’s why a quadratic can have two answers, one, or none.
This chapter gives you two reliable ways to solve them — factorisation and the famous quadratic formula — and a single number, the discriminant, that tells you in advance how many real answers to expect. Best of all, the formula isn’t something to blindly memorise: you’ll watch it being built from scratch.
The Big Idea
A quadratic equation is any equation that can be written as ax² + bx + c = 0 with a ≠ 0. Its solutions are called roots — the values of x that make the left side equal zero, exactly the zeroes of the polynomial ax² + bx + c. Because a quadratic polynomial has at most two zeroes, a quadratic equation has at most two roots. You find them either by factorising into two linear pieces, or by the quadratic formula x = (−b ± √(b² − 4ac)) / 2a. The quantity under the root, D = b² − 4ac, the discriminant, decides the whole story: D > 0 → two roots, D = 0 → one repeated root, D < 0 → none.
Let’s Break It Down
What counts as a quadratic equation
A quadratic equation in x has the standard form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0 (if a were 0 there’d be no x² term and it would just be linear). Examples: 2x² + x − 300 = 0, 2x² − 3x + 1 = 0, 1 − x² + 300 = 0.
The catch: many equations look quadratic but aren’t, and some look cubic but secretly are. Always simplify to standard form first before deciding — bring everything to one side and collect like terms.
Decide whether each is a quadratic equation: (i) x(x + 1) + 8 = (x + 2)(x − 2); (ii) (x + 2)³ = x³ − 4.
- (i) Expand both sides: LHS = x² + x + 8; RHS = x² − 4.
- Bring to one side: x² + x + 8 = x² − 4 → x + 12 = 0. The x² terms cancelled! Degree is 1, so this is not a quadratic equation.
- (ii) Expand LHS: (x + 2)³ = x³ + 6x² + 12x + 8. Set equal to x³ − 4.
- x³ + 6x² + 12x + 8 = x³ − 4 → the x³ cancels → 6x² + 12x + 12 = 0, i.e. x² + 2x + 2 = 0. This is a quadratic equation (it only looked cubic). So always simplify before judging.
Roots: what we’re hunting for
A real number α is a root of ax² + bx + c = 0 if putting x = α makes it true, i.e. aα² + bα + c = 0. “Root”, “solution”, and “zero of the polynomial” all mean the same thing here. For example, x = 1 is a root of 2x² − 3x + 1 = 0 because 2(1)² − 3(1) + 1 = 0. Since a quadratic polynomial has at most two zeroes, a quadratic equation has at most two roots — keep that ceiling in mind.
Solving by factorisation (splitting the middle term)
The plan rests on one fact: if a product of two things is zero, at least one of them is zero. So if we can write ax² + bx + c as a product of two linear factors, say (px + q)(rx + s), then (px + q)(rx + s) = 0 forces px + q = 0 or rx + s = 0 — two small equations we can each solve.
To factorise, split the middle term bx into two parts whose coefficients multiply to a × c and add to b.
Find the roots of 2x² − 5x + 3 = 0.
- Here a = 2, b = −5, c = 3. We need two numbers that multiply to a × c = 2 × 3 = 6 and add to b = −5. Those are −2 and −3 (since −2 × −3 = 6 and −2 + −3 = −5).
- Split the middle term: 2x² − 2x − 3x + 3 = 0.
- Group and take common factors: 2x(x − 1) − 3(x − 1) = 0 → (2x − 3)(x − 1) = 0.
- Set each factor to zero: 2x − 3 = 0 → x = 3/2; and x − 1 = 0 → x = 1. So the roots are x = 3/2 and x = 1. (Check: 2(1)² − 5(1) + 3 = 0 ✓.)
The prayer hall has area 300 m² with length one metre more than twice the breadth: solve 2x² + x − 300 = 0 for the breadth x.
- a = 2, b = 1, c = −300. Need two numbers multiplying to a × c = 2 × (−300) = −600 and adding to 1. Those are 25 and −24.
- Split: 2x² + 25x − 24x − 300 = 0.
- Group: x(2x + 25) − 12(2x + 25) = 0 → (2x + 25)(x − 12) = 0.
- So x = −25/2 = −12.5 or x = 12. Breadth can’t be negative, so breadth = 12 m and length = 2x + 1 = 25 m.
When you factorise a quadratic and one factor repeats — like (3x − √2)² = 0 — how many distinct roots are there, and what do we call them?
Deriving the quadratic formula by completing the square
Factorisation is quick when nice factors exist — but they often don’t. We need a method that always works. The idea is completing the square: reshape ax² + bx + c so the x-part becomes a perfect square like (x + something)², which we can then undo with a square root.
This derivation is the proof of the quadratic formula. Here it is, step by step.
Theorem (Quadratic Formula). The roots of ax² + bx + c = 0 (a ≠ 0) are x = (−b ± √(b² − 4ac)) / 2a, provided b² − 4ac ≥ 0.
Proof. Start from ax² + bx + c = 0. Since a ≠ 0, divide every term by a:
x² + (b/a)x + c/a = 0.
Now look at x² + (b/a)x. We want to turn this into a perfect square. Recall (x + k)² = x² + 2kx + k². Matching 2k with b/a gives k = b/(2a), so we need to add k² = b²/(4a²). Add and subtract it (adding zero, so the equation is unchanged):
x² + (b/a)x + b²/(4a²) − b²/(4a²) + c/a = 0.
The first three terms are now a perfect square:
(x + b/(2a))² = b²/(4a²) − c/a.
Combine the right side over the common denominator 4a²: b²/(4a²) − c/a = (b² − 4ac) / (4a²). So:
(x + b/(2a))² = (b² − 4ac) / (4a²).
If b² − 4ac ≥ 0, take the square root of both sides (the right side’s denominator 4a² = (2a)² is a perfect square):
x + b/(2a) = ± √(b² − 4ac) / (2a).
Finally subtract b/(2a) and combine:
x = −b/(2a) ± √(b² − 4ac) / (2a) = (−b ± √(b² − 4ac)) / 2a.
That is the formula. ∎ This single expression solves every quadratic with real roots — credited to Sridharacharya (around 1025 CE), who derived it by exactly this completing-the-square method.
Solve x² + 7x − 60 = 0 using the quadratic formula.
- Identify a = 1, b = 7, c = −60.
- Compute the discriminant: b² − 4ac = 7² − 4(1)(−60) = 49 + 240 = 289.
- Since 289 ≥ 0, apply x = (−b ± √(b² − 4ac)) / 2a = (−7 ± √289) / 2 = (−7 ± 17) / 2.
- So x = (−7 + 17)/2 = 5 or x = (−7 − 17)/2 = −12. The roots are x = 5 and x = −12.
The discriminant and the nature of the roots
Look again at the formula: everything hinges on b² − 4ac, the part under the square root. We call it the discriminant, written D = b² − 4ac. It “discriminates” between the three possible situations before you finish solving — because you can’t take the real square root of a negative number.
Theorem (Nature of roots). For ax² + bx + c = 0 (a ≠ 0) with D = b² − 4ac:
Proof. From the formula, x = (−b ± √D) / 2a.
- If D > 0: √D is a positive real number, so −b + √D and −b − √D are different. The two values (−b + √D)/2a and (−b − √D)/2a are distinct real roots.
- If D = 0: √D = 0, so x = −b/2a ”± 0” gives the same value twice. There is one value, counted as two equal real roots (a repeated root) at x = −b/2a.
- If D < 0: there is no real number whose square is the negative D, so √D is not real. Hence the formula yields no real roots. ∎
Geometrically, the graph of y = ax² + bx + c is a parabola, and the real roots are exactly where it crosses the x-axis. The three cases are simply the three ways a parabola can sit relative to that axis:
| Discriminant D = b² − 4ac | Nature of roots | Parabola vs x-axis |
|---|---|---|
| D > 0 | two distinct real roots | cuts the axis at two points |
| D = 0 | two equal real roots (one repeated) | just touches the axis |
| D < 0 | no real roots | never meets the axis |
Find the discriminant of 2x² − 4x + 3 = 0 and state the nature of its roots.
- Here a = 2, b = −4, c = 3.
- D = b² − 4ac = (−4)² − 4(2)(3) = 16 − 24 = −8.
- Since D = −8 < 0, there is no real square root, so the equation has no real roots.
Find the discriminant of 3x² − 2x + 1/3 = 0, find the nature of its roots, and find them if real.
- Here a = 3, b = −2, c = 1/3.
- D = (−2)² − 4 × 3 × (1/3) = 4 − 4 = 0. So there are two equal real roots.
- The repeated root is x = −b/2a = −(−2)/(2 × 3) = 2/6 = 1/3.
- So the roots are x = 1/3 and x = 1/3 (equal).
Quadratics in word problems
Many real situations land on a quadratic. The recipe is the same every time: name the unknown, translate the words into ax² + bx + c = 0, solve, then reject any answer that doesn’t fit the situation (a length, age, or count can’t be negative).
A pole is to be erected on the boundary of a circular park of diameter 13 m so that its distances from two diametrically opposite gates A and B differ by 7 m. Is it possible? If so, find the distances from each gate.
- Let BP = x m. Since the distances differ by 7, AP = (x + 7) m. As AB = 13 m is a diameter, angle APB = 90°, so by Pythagoras AP² + BP² = AB².
- (x + 7)² + x² = 13² → x² + 14x + 49 + x² = 169 → 2x² + 14x − 120 = 0 → divide by 2: x² + 7x − 60 = 0.
- Discriminant D = 7² − 4(1)(−60) = 49 + 240 = 289 > 0, so real roots exist — yes, the pole can be placed.
- x = (−7 ± √289)/2 = (−7 ± 17)/2 → x = 5 or x = −12. Distance can’t be negative, so x = 5. The pole is 5 m from gate B and 12 m from gate A.
Common Mistakes
ax² + bx + c = 0 is quadratic as long as it has an x² written in it.
The x² term jumps out visually, so spotting it feels like enough — and in a tidy textbook equation it usually is.
It is quadratic only if, after simplifying to standard form, the coefficient of x² is non-zero (a ≠ 0). In (x + 2)(x − 2) = x² − 4 the x² cancels and it's linear; in (x + 2)³ = x³ − 4 the x³ cancels and a cubic-looking equation becomes quadratic. Always simplify first.
To solve a quadratic you can divide both sides by x to cancel it.
Cancelling worked all through earlier algebra, so dividing by x looks like a fair, familiar move.
Dividing by x silently assumes x ≠ 0 and throws away the root x = 0. For x² − 5x = 0, factor as x(x − 5) = 0 to keep both roots x = 0 and x = 5. Never divide an equation by the variable.
If a product equals any number, say (x − 2)(x − 3) = 6, then x − 2 = 6 or x − 3 = 6.
The 'set each factor equal' rule is so handy that it feels like it should work for any right-hand value, not just zero.
The zero-product rule needs ZERO on one side: only 'product = 0' forces a factor to be 0. First expand and rearrange (x − 2)(x − 3) = 6 to x² − 5x + 6 − 6 = 0 → x² − 5x = 0, THEN factorise and set factors to zero.
If the discriminant is negative, the quadratic equation has no solution at all, so the problem is unsolvable.
In Class 10 'no real roots' and 'no solution' sound identical, so a negative D feels like a dead end or a mistake.
D < 0 means no REAL roots — the parabola simply never touches the x-axis. That is a perfectly valid answer (e.g. 'no, such a rectangle can't exist'). It signals impossibility within real numbers, not an error in your working.
Quick Check
Which of these is a quadratic equation in standard form ax² + bx + c = 0 with a ≠ 0?
To solve 6x² − x − 2 = 0 by splitting the middle term, which two numbers do you use (they multiply to a × c and add to b)?
The discriminant of a quadratic equation is D = 0. What is the nature of its roots?
In the quadratic formula, what is the expression under the square root sign?
Practice Problems
Easy
Find the roots of x² − 3x − 10 = 0 by factorisation.
We need two numbers multiplying to a × c = 1 × (−10) = −10 and adding to b = −3. Those are −5 and 2.
Split: x² − 5x + 2x − 10 = 0 → x(x − 5) + 2(x − 5) = 0 → (x − 5)(x + 2) = 0.
So x − 5 = 0 or x + 2 = 0, giving x = 5 and x = −2. (Check: 5² − 3(5) − 10 = 25 − 15 − 10 = 0 ✓.)
Find the discriminant of 2x² − 3x + 5 = 0 and state the nature of its roots.
Here a = 2, b = −3, c = 5.
D = b² − 4ac = (−3)² − 4(2)(5) = 9 − 40 = −31.
Since D = −31 < 0, the equation has no real roots.
Medium
Find two consecutive positive integers, the sum of whose squares is 365.
Let the integers be x and x + 1. Then x² + (x + 1)² = 365.
Expand: x² + x² + 2x + 1 = 365 → 2x² + 2x − 364 = 0 → divide by 2 → x² + x − 182 = 0.
Factorise: need numbers multiplying to −182 adding to 1 → 14 and −13. So (x + 14)(x − 13) = 0 → x = −14 or x = 13.
The integers are positive, so x = 13. The numbers are 13 and 14. (Check: 13² + 14² = 169 + 196 = 365 ✓.)
Solve 3x² − 2√6·x + 2 = 0 by factorisation, and state the nature of its roots.
Split the middle term −2√6·x. We need parts multiplying to (3)(2) = 6 — written with √6, these are −√6·x and −√6·x (since √6 × √6 = 6 and they add to −2√6).
3x² − √6·x − √6·x + 2 = 0 → √3·x(√3·x − √2) − √2(√3·x − √2) = 0 → (√3·x − √2)(√3·x − √2) = 0.
So the factor (√3·x − √2) repeats: √3·x − √2 = 0 → x = √2/√3.
Both roots are equal: x = √2/√3 and x = √2/√3. (Indeed D = (2√6)² − 4(3)(2) = 24 − 24 = 0, confirming two equal real roots.)
Challenge
A train travels 480 km at a uniform speed. If its speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Let the speed be x km/h. Time = distance ÷ speed = 480/x hours.
At the slower speed (x − 8), time = 480/(x − 8), which is 3 hours more:
480/(x − 8) = 480/x + 3.
Multiply through by x(x − 8): 480x = 480(x − 8) + 3x(x − 8).
480x = 480x − 3840 + 3x² − 24x → 0 = 3x² − 24x − 3840 → divide by 3 → x² − 8x − 1280 = 0.
Factorise (need numbers multiplying to −1280, adding to −8 → 32 and −40): (x − 40)(x + 32) = 0 → x = 40 or x = −32.
Speed can’t be negative, so the train’s speed is 40 km/h. (Check: 480/40 = 12 h; 480/32 = 15 h; difference 3 h ✓.)
Find the value of k for which the quadratic equation 2x² + kx + 3 = 0 has two equal roots.
Two equal roots means the discriminant is zero: D = b² − 4ac = 0.
Here a = 2, b = k, c = 3, so D = k² − 4(2)(3) = k² − 24.
Set D = 0: k² − 24 = 0 → k² = 24 → k = ±√24 = ±2√6.
So k = 2√6 or k = −2√6. (For either value the equation has two equal real roots.)
Summary
You should now be able to explain:
- A quadratic equation is one that simplifies to ax² + bx + c = 0 with a, b, c real and a ≠ 0. Always simplify to standard form before deciding if an equation is quadratic.
- A root (= solution = zero of the polynomial) is a value α with aα² + bα + c = 0. A quadratic has at most two roots.
- Factorisation: split the middle term bx into parts that multiply to a × c and add to b, factor into two linears, then set each factor to zero (this needs zero on one side).
- Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a, derived by completing the square — it always works when real roots exist.
- The discriminant D = b² − 4ac gives the nature of the roots: D > 0 → two distinct real roots; D = 0 → two equal real roots; D < 0 → no real roots. These match a parabola cutting, touching, or missing the x-axis.
- In word problems, translate to a quadratic, solve, and reject answers (negative lengths, ages, speeds) that don’t fit the situation.
What’s Next
Quadratics were about one squared unknown. Next, in Arithmetic Progressions, we step back to patterns of numbers that grow by a fixed step — like 5, 8, 11, 14, … — and find neat formulas for the nth term and the sum of many terms. (And watch for a surprise: adding up an AP often leads straight back to a quadratic equation.)