Quadratic Equations
A quadratic equation contains an equal sign and an unknown raised to the power 2. For example: 2x2 – 5x – 3 = 0
n2 + 50 = 27n
0 = (4a – 9)(2a + 1)
49 = k2
Are all quadratic equations. Discussion: can you see why
0 = (4a – 9)(2a + 1) is a quadratic equation?
One of the main objectives of the chapter is to find ways of solving quadratic equations,
i.e. finding the value(s) of the unknown that make the equation true.
Solving Quadratic Equations
One way of solving quadratic equation is to apply the following argument to a quadratic expression that has been factorized.
If the product of two numbers is 0, then one of the numbers (or possibly both of them) must be 0. For example,
3 × 0 = 0, 0 × 5 = 0 and 0 × 0 = 0
In general, if a × b = 0
Then either a = 0
Or b = 0
Or both a and b are 0
Example 1
Solve the equation (x – 2)(x + 7) = 0. If (x – 2)(x + 7) = 0
Then either x – 2 = 0 or x + 7 = 0
x = 2 or -7
Example 2
Solve the equation d(d – 4)(d + 62) = 0.
(3a + 2)(2a – 7) = 0, then any one of the four factors of the LHS may be 0,
i.e d = 0 or d – 4 = 0 or d + 6 = 0 twice.
⟹ d = 0, 4 or -6 twice.
EVALUATION
Solve the following equations. 1. 3d2(d – 7) = 0
2. (6 – n)(4 + n) = 0
3. A(2 – a)2(1 + a) = 0
Solving quadratic equations using factorization method
The LHS of the quadratic equation m2 – 5m – 14 = 0 factorises to give (m + 2)(m – 7) = 0.
Example 1
Solve the equation 4y2 + 5y – 21 = 0 4y2 + 5y – 21 = 0
⟹ (y + 3)(4y – 7) = 0
⟹either y + 3 = 0 or 4y – 7 = 0 y = – 3 or 4y = 7
y = – 3 or y = 7/4
y = -3 or 13
4
check: by substitution:
if y = -3
4y2 + 5y – 21 = 36 – 15 – 21 = 0
If y = 13,
4
4y2 + 5y – 21 = 4 x 7/4 x 7/4 + 5 x 7/4 – 21
= 49 + 35 – 21 = 0
4 4
Example 2
Solve the equation m2 = 16 Rearrange the equation.
If m2 = 16
Then m2 – 16 = 0
Factorise (difference of two squares) (m – 4)(m + 4) = 0
Either m – 4 = 0 or m + 4 = 0 m = +4 or m = -4
m = ±4
EVALUATION
Solve the following quadratic equations:
1. h2 – 15h + 54 = 0
2. 12y2 + y – 35 = 0
3. 4a2 – 15a = 4
4. v2 + 2v – 35 = 0
GENERAL EVALUATION
Solve the following equations:
1. y2(3 + y) = 0
2. x2(x + 5)(x – 5) = 0
3. (v – 7)(v – 5)(v – 3) = 0
4. 9f2 + 12f + 4 = 0
WEEKEND ASSIGNMENT
Solve the following equations. Check the results by substitution. 1. (4b – 12)(b – 5) = 0 A. ½, 4 B. 3, 5 C. 4, 6 D.5, 3
2. (11 – 4x)2 = 0 A.11, 3 B.23, 3 C. 23 twice D. 24 twice
3 4 4 3
3. (d – 5)(3d – 2) = 0 A. 5,2
3
B. 4, 5 C. 5, 9 D. 2, 5
3
Solve the following quadratic equations
4. u2 – 8u – 9 = 0A. – 9, 1 B. -1, 9 C. 1, 8 D. 9 , -1
5. c2 = 25 A. 5 B. -5 C.+5 D.±5
THEORY
Solve the equation 1. 2x2 = 3x + 5
2. a2 – 3a = 0
3. p2 + 7p + 12 = 0
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