

Quadratic Equations and Quadratic
Function 
Quadratic equation 
Solving quadratic equations by completing the
square, the quadratic formula 
Solving quadratic equations by factoring,
Vieta's formula 





Quadratic equation 
A
quadratic equation is a polynomial equation of the second
degree. The general or standard form is 
ax^{2}
+ bx + c = 0,
a is
not 0 
where a,
b
and c
are coefficients. 

Solving quadratic equations by completing the
square, the quadratic formula 
The solutions of the equation
are called roots (which may be real or complex) and are given by
the quadratic formula. 
We
use the completing the square method to derive the quadratic
formula to find the roots, 

The
roots can also be written 

If
c = 0,
the roots of the
equation ax^{2}
+ bx = 0
we can obtain by factoring 
x(ax
+ b) = 0 
or
by using the quadratic formula, so we get x_{1}=
0 and x_{2
}= b/a. 

Solving quadratic equations by factoring, Vieta’s formula 
A
quadratic trinomial ax^{2
}+ bx^{
}+ c
can be factorized as 
ax^{2
}+ bx^{
}+ c
= a·[x^{2
}+ (b/a)·x
+ c/a]
= a·(x

x_{1})(x

x_{2}), 
where
x_{1}
+ x_{2}
= b/a and
x_{1}·
x_{2}
= c/a 

That
means, to factor a quadratic trinomial we should find such a
pair of numbers x_{1}
and x_{2}
whose sum equals b/a
and whose product equals c/a. 
So,
if a >
0 and the constant term c
negative, then the signs of x_{1}
and x_{2}
will be different while, when c
is positive, their signs will be the same. 
Then we use zero product principle to solve the quadratic equation,
that is, a product a
· b = 0
if at least one of its factors is 0. 
Examples:
Find the roots of each
quadratic equation by factoring:

a) x^{2
}  3x
10
= 0

since
x^{2
}  3x
10
= x^{2
} + (5
+
2)·x
+ (5)·(+2)
= x^{2
}  5x
+
2x
10

= x ·
(x
 5)
+ 2 ·
(x
 5)
= (x
 5)
·
(x
+ 2),

then
x
 5
= 0
=>
x_{1} =
5,
or x
+ 2
= 0
=>
x_{2} =
2. 

b) 2x^{2
} 7x
+ 3
= 0 
as 2x^{2
} 7x
+ 3
= 2[x^{2
}  (7/2)x
+ 3/2]
= 2[x^{2
}  (1/2)x
 3x
+ 3/2]
= 
= 2[x(x
 1/2)
 3(x  1/2)]
= 2
(x  1/2)(x
 3)
= (2x
 1)
(x  3), 
then
2x
 1
= 0
=>
x_{1} =
1/2,
or x  3
= 0
=>
x_{2} =
3 

c) 3x^{2
} x
 2
= 0 
as 3x^{2
} x
 2
= 3[x^{2
}  (1/3)x
 2/3]
= 3[x^{2} +
(2/3)x
 x
 2/3]
= 
= 3[x(x
+ 2/3)

(x + 2/3)]
= 3·(x
+ 2/3)(x
 1)
= (3x + 2)(x
 1), 
then
3x + 2
= 0
=>
x_{1} =
2/3,
or x
 1
= 0
=>
x_{2} =
1. 


Example:
Given
are leading
coefficient a_{2}_{
} =
1
and
the pair
of conjugate complex
roots, 
x_{1 } =
1 +
i
and
x_{2
} =
1  i,
of
a second
degree polynomial; 
a)
find
the polynomial using
the above theorem, 
b)
make a check of the solutions of the polynomial using the
quadratic formula. 
Solution: 
a)
By
plugging the given values into a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{2}(x
 x_{1})
(x
 x_{2}) 

a_{2}x^{2}
+ a_{1}x
+ a_{0} = 1[x

(1
+ i)]
· [x

(1  i)]
= [(x
 1)
 i]
· [(x  1)
+ i]
=


= [(x  1)^{2}
 i^{2}]
=
(x^{2}
 2x
+ 1 +
1)
=
 x^{2}
+ 2x
 2 

b)
Check the polynomial for the roots, 











Precalculus
contents B 



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