CHAPTER 7. SOLVINGQUADRATIC EQUATIONS 7.4
Consider the general form of the quadratic function:
f(x) = ax^2 + bx + c.Factor out the a to get:
f(x) = a�
x^2 +
b
ax +
c
a�
. (7.2)
Now we need to do some detective work to figure out how to turn (7.2) into a perfect square plus some
extra terms. We know that for a perfect square:
(m + n)^2 = m^2 + 2mn + n^2and
(m− n)^2 = m^2 − 2 mn + n^2
The key is the middle term on the right hand side, which is 2 × the first term× the second term of the
left hand side. In (7.2),we know that the first term is x so 2 × the second term isba. This means that
the second term is 2 ba. So,
�
x +
b
2 a� 2
= x^2 + 2b
2 a
x +�
b
2 a� 2
.
In general if you add a quantity and subtract thesame quantity, nothing has changed. This meansif we
add and subtract
�b
2 a� 2
from the right hand sideof Equation (7.2) we will get:f(x) = a�
x^2 +
b
ax +
c
a�
(7.3)
= a�
x^2 +
b
a
x +�
b
2 a� 2
−
�
b
2 a� 2
+
c
a�
(7.4)
= a��
x +�
b
2 a�� 2
−
�
b
2 a� 2
+
c
a�
(7.5)
= a�
x +�
b
2 a�� 2
−
b^2
4 a+ c (7.6)We set f(x) = 0 to find its roots, whichyields:
a�
x +
b
2 a� 2
=
b^2
4 a− c (7.7)Now dividing by a and taking the square root of both sides gives the expression
x +
b
2 a=±
�
b^2
4 a^2−
c
a(7.8)
Finally, solving for x implies that
x =−b
2 a±
�
b^2
4 a^2−
c
a=−
b
2 a±
�
b^2 − 4 ac
4 a^2which can be further simplified to:
x =
−b±√
b^2 − 4 ac
2 a(7.9)
These are the solutionsto the quadratic equation. Notice that there aretwo solutions in general, but
these may not always exists (depending on the sign of the expression b^2 − 4 ac under the square root).
These solutions are alsocalled the roots of the quadratic equation.