Chapter 4: Quadratic Equation
Graphical solution of quadratic equations, with additional source pages on linear programming and circle equations — reconstructed from the supplied 10-page handwritten PDF.
Original Scanned PDF – View Notes
Linear Programming Problem
Optimize
Subject to:
5x + 10y ≤ 50
x + y ≥ 1
y ≤ 4
x ≥ 0, y ≥ 0
Boundary Equations
5x + 10y = 50
x + y = 1
y = 4
x = 0
y = 0
Testing the Origin
For 5x + 10y ≤ 50:
5(0) + 10(0) ≤ 50
0 ≤ 50 — True
Therefore, the boundary side containing the origin satisfies this inequality.
For x + y ≥ 1:
0 + 0 ≥ 1
0 ≥ 1 — False
Therefore, the feasible side is the side opposite the origin.
Feasible Region
The handwritten graph identifies the feasible-region vertices as:
| Point | x | y | Z = 2x+y |
|---|---|---|---|
| A | 0 | 1 | 1 |
| B | 1 | 0 | 2 |
| C | 10 | 0 | 20 |
| D | 2 | 4 | 8 |
| E | 0 | 4 | 4 |
Maximum value: Z = 20 at (10,0)
Minimum value: Z = 1 at (0,1)
Straight-Line Equation from Given Points
The source gives three points approximately as:
Equation of AB
Using the two-point form:
(y−2) / (x−3) = (6−2) / (−3−3)
y−2 = (−4/6)(x−3)
After simplification, the handwritten source gives:
2x + 3y = 12
Quadratic Equations Listed in the Source
(i) x² − x − 2 = 0
(ii) x² + x − 2 = 0
(iii) x² − 3x + 2 = 0
(iv) x² − 5x + 6 = 0
(v) x² − x − 12 = 0
(vi) x² − 4x + 3 = 0
(vii) x² − 3x − … = 0
Graphical Solution of Quadratic Equations
Example 1: x² − 2x − 3 = 0
Rewrite the quadratic equation as:
Therefore, draw the two graphs:
Values for y = x²
| x | 0 | 1 | −1 | 2 | −2 | 3 | −3 | 4 | −4 |
|---|---|---|---|---|---|---|---|---|---|
| y | 0 | 1 | 1 | 4 | 4 | 9 | 9 | 16 | 16 |
Values for y = 2x + 3
| x | 1 | 2 |
|---|---|---|
| y | 5 | 7 |
The graphs intersect at approximately (−1,1) and (3,9).
Therefore, the roots are:
x = −1 and x = 3
Example 2: x² − x − 2 = 0
Rewrite as:
Draw:
Values for y = x+2
| x | 1 | 2 |
|---|---|---|
| y | 3 | 4 |
The corresponding intersections give the roots:
x = −1 and x = 2
Graph of y = x² + 2x − 8
Vertex
For y = ax² + bx + c:
x-coordinate of vertex = −b/(2a)
= −2/(2×1)
= −1
At x=−1:
y=(−1)²+2(−1)−8
=1−2−8
=−9
Therefore, the vertex is (−1,−9).
Value Table Visible in the Source
| x | −2 | −1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|---|
| y | −8 | −9 | −8 | −5 | 0 | 7 | 16 | 27 | 40 |
Circle Equation Problem
The source gives the circle:
One end of a diameter is shown as (0,2), and the other end is (a,b).
Using the Centre of the Circle
Compare with the standard form:
2g = 4 ⇒ g = 2
2f = −6 ⇒ f = −3
Centre = (−g,−f)
∴ Centre = (−2,3)
Using the Midpoint of the Diameter
Midpoint of (0,2) and (a,b) is:
((a+0)/2, (b+2)/2)
This equals the centre (−2,3).
a/2 = −2 ⇒ a = −4
(b+2)/2 = 3 ⇒ b+2=6 ⇒ b = 4
Therefore, the other end of the diameter is:
(−4,4)
Discussion
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