Deriving the Quadratic Formula
A Factorization Route
The quadratic formula for finding roots of expressions comes follow
from [i] completing the square and then [ii] factoring if [i] results in
the difference of two squares. Numerical examples were given in the
previous lesson. Here is an algebraic approach.
Alternative Expositions
For skill development and perfection, preparation for calculus, you
should follow both the equation solving and factorization routes
What is a reducible quadratic?
Definition: A quadratic polynomial ax2+bx+c is
irreducible with respect to the real numbers if and only
if there are no real real numbers r and s such that
ax2+bx+c = a[x-r][x-s].
Example of an irreducible quadratic: 5[[x-2]2 +
16] > 5*16 = 80 > 0 has no zeroes because all its values are
> 80 > 0. If [x-2]2 + 16 were
equal to [x-r][x-s] for one or two real numbers r and s, it would be
equal to zero when x = r or x = s. That possibility is inconsistent with
the observation that 5[[x-2]2 + 16] > 0 for all real
numbers x. Consistency requires that [[x-2]2 + 16] be
irreducible.
Definition: A quadratic polynomial ax2+bx+c is
reducible with respect to the real numbers if and only if
ax2+bx+c = a[x-r][x-s] for some real numbers r and s.
Example of a reducible quadratic:
5[[x-2]2 -9]
= 5[[x-2]2 -32]
= 5[[x-2+3][x-2-3]]
= 5[x+1][x-5]
= 5[x-r][x-s] if r = -1 and s = 5.
Thus 5[[x-2]2 -9] = 5[x2-4x+5] =
5x2-20x+25 is reducible.
Theorem A: If a quadratic ax 2+bx+c =
a[[x-h] 2 + k ] for some real numbers h and k with k > 0
then the quadratic is irreducible with respect to the real numbers. It
has no real roots.
Proof: The absolute value of the quadratic
|ax2+bx+c| = |a[[x-h]2 + k ]| =
|a|.|[x-h]2 + k| >
|a|.k > 0
can is non-zero for all real numbers x. So the quadratic cannot have
the value zero.
Derivation of Quadratic Formula, Step I
Theorem B: If a quadratic ax 2+bx+c =
a[[x-h] 2 + k ] for some real numbers h and k with k
< 0 then this quadratic is reducible and its has one or
two real roots, namely
\[ x = h+\sqrt{-k} \hbox{ and } x = h-\sqrt{-k} \]
These formulas for roots both give x = h when k = 0.
Proof -Equation Solving Route: If k < 0, then
[x-h] 2 + k = 0 when and only when [x-h] 2 = -k
The latter is equivalent to: \[x-h = +\sqrt k \hbox{ or }x-h = -\sqrt k\]
The latter in turn are equivalent to \[x = h +\sqrt k = r \hbox{ or }x =
h- \sqrt k =s\]
This gives the first way to solve a[[x-h]2 + k ] = 0 or
ax2+bx+c = 0 when ax2+bx+c =
a[[x-h]2 + k ]. The roots or solutions are equidistant from
the axis of symmetry, the line x = h.
Proof - difference of two squares route: If k = 0, then
ax2+bx+c = a[x-h]2 is zero when and only
when x = h. In the remaining case k < 0. So -k is positive and
we can write -k as the square of its principal square root \[ w =
\sqrt{-k} \] Hence -k = w2.
The foregoing gives
ax2+bx+c
= a[[x-h]2 + k ]
= a[[x-h]2 - [-k] ] [algebraic trick: +k =
-[-k]
adding k is the same as
subtracting -k
= a[[x-h]2 - w2 ]
since -k = w2
= a[x-h+w][x-h-w] due to difference of 2 squares.
= a[x - r][x - s]
where $r = h + w = h +\sqrt {-k} $ and $s = h +\sqrt {-k} $ give two roots or
zeroes of the quadratic ax2+bx+c .
Above 5[[x-2]2 -9] = a[[x-h]2 + k ] when a = 5, h =
2 and k = -9. The above theorem or its proof puts or takes $w = \sqrt{-k} = \sqrt 9
= 3,$ and so implies
Derivation of Quadratic Formula, Step II
Conversion of ax2+bx+c into the form
a[[x-h]2 + k] leads to expressions for h and k in terms of a,
b and c. The substitution of those expressions into
\[x = h +\sqrt {-k} \hbox{ or }x = h- \sqrt{-k} \]
provides the quadratic formula. The second proof, a difference of two squares,
route also shows how to express quadratic expressions
ax2+bx+c as a product of factors.
Theorem C: Each quadratic expression
ax2+bx+c = a[[x-h]2 + k]
where
\[ h = -\frac b{2a} \mbox{ and } -k = \frac{b^2-4ac}{(2a)^2} \]
provided the x2-coefficient $a \ne 0$
Proof: The proof here follows the path taken in the proof of
Theorem B.
\begin{eqnarray*}
ax^2+bx+c &=& a\left[x^2+\frac ba x + \frac ca \right]
\\ &=& a\left[ \left( x+\frac b{2a}\right)^2 - \frac{b^2}{(2a)^2} + \frac ca \right]
\\ &=& a\left[ \left( x+\frac b{2a}\right)^2 - \frac{b^2}{4a^2} + \frac {4ac}{4a^2} \right]
\\ &=& a\left[ \left( x+\frac b{2a}\right)^2 + \frac {4ac-b^2}{4a^2} \right]
\\ &=& a\left[ \left( x+\frac b{2a}\right)^2 - \frac{b^2-4ac}{4a^2} \right]
\\ &=& a[(x+h)^2+k ]
\end{eqnarray*}
where
\[ h = -\frac b{2a} \mbox{ and } k = -\frac{b^2-4ac}{(2a)^2} \]
Derivation of Quadratic Formula, Step III
Recall
Theorem B: If a quadratic ax 2+bx+c =
a[[x-h] 2 + k ] for some real numbers h and k with k
< 0 then this quadratic is reducible and its has one or
two real roots, namely
\[ x = h+\sqrt{-k} \hbox{ and } x = h-\sqrt{-k} \]
For k zero or negative, the formula
ax2+bx+c = a[[x-h]2 + k]
where
\[ h = -\frac b{2a} \mbox{ and } -k = \frac{b^2-4ac}{(2a)^2} \]
The latter expression gives two roots of the quadratic equation $ax^2+bx+c =0,$ namely
\begin{eqnarray*} x &=& h\pm\sqrt{-k}
\\ &=& -\frac b{2a} \pm \sqrt{\frac{b^2-4ac}{(2a)^2}}
\\ &=& -\frac b{2a} \pm \frac{\sqrt{b^2-4ac}}{ 2a }
\\ &=& \frac{-b \pm\sqrt{b^2-4ac}}{ 2a }
\end{eqnarray*}
The last expression is called the quadratic formula.
The first square in C2 - R2 is
Remarks and Conclusions
The factorization
|
ax2+bx+c = a [
|
x +
|
b + sqrt[b2 -4ac]
2a
|
]
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[
|
x +
|
b - sqrt[b2 -4ac]
2a
|
]
|
says
ax2+bx+c = = a[x -r][x-s]
where
|
r = -
|
b + sqrt[b2 -4ac]
2a
|
|
-b - sqrt[b2 -4ac]
2a
|
and
|
s = -
|
b - sqrt[b2 -4ac]
2a
|
|
-b + sqrt[b2 -4ac]
2a
|
Observe ½[r+s] = - b/2a = h
Thus the two roots s and r are given by or can be summarized by the
formulas
|
x =
|
-b ±
sqrt[b2 -4ac]
|
|
|
2a
|
Here the plus sign gives the s-value for x and the negative sign gives
the r-value for x. The quadratic formula with the plus-minus
sign ± is actually two
formulas in one.
Discriminant d = b2 - 4ac and Counting Roots
A. If the discriminant d = b2 - 4ac < 0
then k = [4ac-b2]/[4a2] > 0 and the
y = ax2+bx+c = a[[x-h]2 + k ]
has magnitude > |ak| > 0 and so no real roots. Here
completing the square would lead to the sum of squares and not a
difference.
B. If the discriminant d = b2 - 4ac = 0
then k = [4ac-b2]/[4a2] = 0 and the
y = ax2+bx+c = a[x-h]2
has one real root y = 0 at x = h = -b/2a on the axes of
symmetry x = h = -b/2a. Here completing the square would lead
to zero value for k, and neither a difference nor a sum of squares.
C. If the discriminant d = b2 - 4ac > 0
then k = [4ac-b2]/[4a2] < 0 and the
y = ax2+bx+c = a[[x-h]2 + k ]
has two real roots r and s with $\frac{r+s}2 = - \frac b{2a} = h.$ Here
completing the square would lead to the difference of two squares.
The quadratic formula $x=\frac{-b+\pm\sqrt{b^2-4ac}}{ 2a } $ gives two
roots, namely
\[ r= \frac{-b+ \sqrt{b^2-4ac}}{ 2a } \]
and
\[ s= \frac{-b -\sqrt{b^2-4ac}}{ 2a } \]
with
ax2+bx+c = a[x -r][x-s]
The property $\frac{r+s}2 = - \frac b{2a} = h $ implies the axis of symmetry x = h =
-b/2a is midway between the two roots or zeroes of the quadratic
ax2+bx+c = a[x -r][x-s]
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