Standard form, methods of solving, nature of roots, and real-life applications
A quadratic equation in the variable x is an equation of the form:
ax² + bx + c = 0
where a, b, c are real numbers and a ≠ 0. The value of x that satisfies the equation is called a root or solution.
If a rectangular garden has length 5 m more than its width, and its area is 150 m², then if width = x: x(x+5) = 150, which gives x² + 5x – 150 = 0. This is a quadratic equation! Solving it gives the actual dimensions of the garden.
In this method, we factorise the quadratic expression and use the zero-product rule: if A × B = 0, then A = 0 or B = 0.
1. Write in standard form: ax² + bx + c = 0
2. Find two numbers p and q such that p × q = a×c and p + q = b
3. Rewrite the middle term: ax² + px + qx + c = 0
4. Factor by grouping, then solve using zero-product rule
a×c = 2×3 = 6 | We need p+q = –5 and p×q = 6
p = –2, q = –3 (since –2 + –3 = –5 and –2 × –3 = 6)
2x² – 2x – 3x + 3 = 0
2x(x – 1) – 3(x – 1) = 0
(2x – 3)(x – 1) = 0
So x = 3/2 or x = 1
Roots: x = 3/2 and x = 1
Find p, q: p×q = 6, p+q = 5 → p = 2, q = 3
x² + 2x + 3x + 6 = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
Roots: x = –2 and x = –3
This is a systematic method that works even when factorisation is not obvious. It involves creating a perfect square trinomial.
1. Write ax² + bx + c = 0 → divide everything by a (if a ≠ 1)
2. Move constant to right: x² + (b/a)x = –c/a
3. Add (b/2a)² to both sides
4. Left side becomes a perfect square: (x + b/2a)²
5. Take square root and solve for x
Divide by 2: x² – 7x/2 + 3/2 = 0
Move constant: x² – 7x/2 = –3/2
Add (7/4)² = 49/16 to both sides:
x² – 7x/2 + 49/16 = –3/2 + 49/16 = –24/16 + 49/16 = 25/16
(x – 7/4)² = 25/16
x – 7/4 = ±5/4
x = 7/4 + 5/4 = 12/4 = 3 OR x = 7/4 – 5/4 = 2/4 = 1/2
Roots: x = 3 and x = 1/2
For ax² + bx + c = 0 (where a ≠ 0):
x = [–b ± √(b² – 4ac)] / 2a
This formula always works and is derived by completing the square on the general form.
a = 3, b = –5, c = 2
x = [–(–5) ± √((–5)² – 4×3×2)] / (2×3)
x = [5 ± √(25 – 24)] / 6 = [5 ± √1] / 6 = [5 ± 1] / 6
x = 6/6 = 1 OR x = 4/6 = 2/3
Roots: x = 1 and x = 2/3
D = b² – 4ac
The discriminant tells us the nature of roots without actually solving the equation.
| Value of D | Nature of Roots | Example |
|---|---|---|
| D > 0 | Two distinct real roots | x² – 5x + 6 = 0, D = 1 > 0 |
| D = 0 | Two equal real roots (repeated) | x² – 4x + 4 = 0, D = 0 |
| D < 0 | No real roots (imaginary) | x² + x + 1 = 0, D = –3 < 0 |
D = (–4)² – 4(1)(4) = 16 – 16 = 0
Since D = 0, the roots are equal and real.
x = 4/2 = 2 (repeated root)
Q: The product of two consecutive positive integers is 306. Find the integers.
Let integers be x and x+1.
x(x+1) = 306 → x² + x – 306 = 0
D = 1 + 1224 = 1225 = 35²
x = (–1 ± 35)/2 → x = 17 or x = –18 (rejected, must be positive)
Answer: 17 and 18
Q: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed of the train.
Let speed = x km/h
Time at x: 360/x | Time at (x+5): 360/(x+5)
360/x – 360/(x+5) = 1
360(x+5) – 360x = x(x+5) → 1800 = x² + 5x → x² + 5x – 1800 = 0
D = 25 + 7200 = 7225 = 85²
x = (–5 + 85)/2 = 40
Answer: Speed = 40 km/h
ax² + bx + c = 0 (a ≠ 0)
x = [–b ± √(b²–4ac)] / 2a
D = b² – 4ac
D>0: 2 distinct real roots
D=0: 2 equal roots
D<0: No real roots
Sum = α+β = –b/a
Product = αβ = c/a