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Quadratic Equation
Mind Map
Quadratic
Equation
Identify QE
Example of QE
and non-QE
Forming QE
from Roots
Solving QE
General Form
Value of a, b
and c
Root of QE
3 Methods
Type of Root
x2 – SoRx +
PoR = 0
Substitution
Factorisation
SoR = -b/a
PoR = c/a
Intersection of
Graph
Completing the
Square
A few example
of SoR and PoR
Formula
Difference Between Quadratic Eqaution and
Quadratic Function
Quadratic Equation
In mathematics, a quadratic equation is a
polynomial equation of the second degree. The
general form is
ax 2 + bx + c = 0
where a ≠ 0. (For a = 0, the equation becomes a
linear equation.)
The letters a, b, and c are called coefficients: the
quadratic coefficient a is the coefficient of x2, the
linear coefficient b is the coefficient of x, and c is
the constant coefficient, also called the free term or
constant term.
Example of Quadratic Equation
2 x2 − 5 = 0
1 − 6 x2 = 3
6 x + 3x 2 = 0
x2 = 0
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Quadratic Equation
Solving Quadratic Equation
3 Methods:
z
Factorisation
z
Completing The Square
z
Quadratic Formula
Factorisation
Example
Solve x2 + 5x + 6 = 0.
Answer
x2 + 5x + 6 = (x + 2)(x + 3)
Set this equal to zero:
(x + 2)(x + 3) = 0
Solve each factor:
x + 2 = 0 or x + 3 = 0
x = –2 or x = – 3
The solution of x2 + 5x + 6 = 0 is x = –3, –2
Completing The Square
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Quadratic Equation
Quadratic Formula
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Quadratic Equation
Forming Quadratic Equation from Its Roots
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Quadratic Equation
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Quadratic Equation
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Quadratic Equation
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Quadratic Equation
Nature of Roots of a Quadratic Equation
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The expression b − 4ac in the general formula is called the discriminant of the equation, as it determines
the type of roots that the equation has.
b 2 − 4ac
b 2 − 4ac
b 2 − 4ac
b 2 − 4ac
> 0 ⇔ two real and distinct roots
= 0 ⇔ two real and equal roots
< 0 ⇔ no real roots
≥ 0 ⇔ the roots are real
e.g. 1:
Find the range of values of k for which the equation
2x 2 + 5x + 3 − k = 0 has two real distinct roots.
e.g. 3:
Find the range of values of p for which the equation
x 2 − 2 px + p 2 + 5 p − 6 = 0 has no real roots.
b 2 − 4ac > 0
b 2 − 4ac < 0
(5) 2 − 4(2)(3 − k ) > 0
25 − 24 + 8k > 0
1 + 8k > 0
(−2 p ) 2 − 4(1)( p 2 + 5 p − 6) < 0
8k > −1
−1
k>
8
−20 p < −24
4 p 2 − 4 p 2 − 20 p + 24 < 0
−20 p + 24 < 0
20 p > 24
24
20
6
p>
5
p>
e.g. 2:
2
The roots of 3x + kx + 12 = 0 are equal. Find k.
b 2 − 4ac = 0
(k ) 2 − 4(3)(12) = 0
k 2 − 144 = 0
k 2 = 144
k = ± 144
k = ±12
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Quadratic Equation
e.g. 4:
2 2
Show that the equation a x + 3ax + 2 = 0
always has real roots.
b 2 − 4ac
= (3a ) 2 − 4(a 2 )(2)
= 9 a 2 − 8a 2
= a2
a2 >0 for all values of a. Therefore
b 2 − 4ac > 0
Proven that
a 2 x 2 + 3ax + 2 = 0 always has real roots.
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