

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 function or
the seconddegree polynomial 
Translated form of
quadratic function 
Vertex (maximum/minimum)  coordinates of translation

Roots or zeros of the
function, axis of symmetry and yintercept 





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, 




Quadratic function or
the seconddegree polynomial 
The
polynomial function of the second degree, f
(x)
=
a_{2}x^{2}
+ a_{1}x
+ a_{0}, is called a
quadratic function. 


y
= f (x)
=
a_{2}x^{2}
+ a_{1}x
+ a_{0}
or y

y_{0}
= a_{2}(x

x_{0})^{2}, 


where 

are the coordinates
of translations of the quadratic 

function. By
setting x_{0}
= 0
and
y_{0}
= 0 we obtain
y
=
a_{2}x^{2},
the source quadratic function.

The turning point V(x_{0},
y_{0})
is called the vertex of the parabola.

Note that the coefficients, a_{2}, a_{1}
and a_{0},
of quadratic function, correspond to the coefficients, a, b
and c,
of

quadratic equation, respectively.

The
real zeros of the
quadratic function: 



The above formula is known quadratic formula
that shows the
symmetry
of the roots relative to the axis of

symmetry of the
parabola.


y
= f (x) =
a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{2}(x
 x_{1})(x
 x_{2})
= a_{2}[x^{2}

(x_{1} +
x_{2})x
+
x_{1}x_{2}] 


The graph of a quadratic function is curve called a parabola. The parabola is symmetric with respect to a vertical line
called the axis of symmetry. 
As
the axis of symmetry passes through the vertex of the parabola
its equation is x
= x_{0}. 
Quadratic
function has the yintercept at the
point ( 0,
a_{0 }).


Translated form of
quadratic function 
The
proof that quadratic function f
(x)
= a_{2}x^{2} + a_{1}x
+ a_{0}
is translation of its source or original
f (x)
= a_{2}x^{2} 
1)
Let calculate the
coordinates of translations of quadratic function using the
formulas, 
substitute
n
= 2 in 



then 



2)
To
get the source quadratic function we should plug the coordinates
of translations (with changed signs) 
into the general form
of the quadratic,
i.e., 

after
expanding and reducing obtained is 
y
=
a_{2}x^{2}
the source quadratic function 
3)
Inversely, by plugging the coordinates of translations into the source quadratic function 
y

y_{0}
= a_{2}(x

x_{0})^{2}, 

and
after
expanding and reducing we obtain 
y
=
a_{2}x^{2}
+ a_{1}x
+ a_{0} the quadratic function
in the general form. 








Intermediate
algebra contents 



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