The Quadratic Equation
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The roots to the quadratic polynomial
y
=
a
x
2
+
b
x
+
c
{\displaystyle y=ax^{2}+bx+c}
are easily derived and many people memorized them in high school:
x
=
−
b
±
b
2
−
4
a
c
2
a
.
{\displaystyle {\begin{aligned}x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\end{aligned}}}
Derivation of the Quadratic Equation
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To derive this set
y
=
0
{\displaystyle y=0}
0
=
a
x
2
+
b
x
+
c
0
=
a
(
x
2
+
b
a
x
+
c
a
)
0
=
x
2
+
b
a
x
+
(
b
2
a
)
2
+
c
a
−
(
b
2
a
)
2
0
=
(
x
+
b
2
a
)
2
+
(
c
a
−
(
b
2
a
)
2
)
.
{\displaystyle {\begin{aligned}0&=ax^{2}+bx+c\\0&=a\left(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}\right)\\0&=x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\\0&=\left(x+{\frac {b}{2a}}\right)^{2}+\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right).\end{aligned}}}
Solving for
x
{\displaystyle x}
0
=
(
x
+
b
2
a
)
2
+
(
c
a
−
(
b
2
a
)
2
)
(
x
+
b
2
a
)
2
=
−
(
c
a
−
(
b
2
a
)
2
)
=
b
2
−
4
a
c
(
2
a
)
2
.
{\displaystyle {\begin{aligned}0&=\left(x+{\frac {b}{2a}}\right)^{2}+\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)\\\left(x+{\frac {b}{2a}}\right)^{2}&=-\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)\\&={\frac {b^{2}-4ac}{\left(2a\right)^{2}}}.\end{aligned}}}
Taking the square root of both sides and putting everything over a common denominator gives
x
+
b
2
a
=
±
b
2
−
4
a
c
(
2
a
)
2
x
=
−
b
±
b
2
−
4
a
c
2
a
.
{\displaystyle {\begin{aligned}x+{\frac {b}{2a}}&=\pm {\sqrt {\frac {b^{2}-4ac}{\left(2a\right)^{2}}}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\end{aligned}}}
Middlebrook has pointed out that this is a poor expression from a numerical point of view for certain values of
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
[Give an example here]
Middlebrook showed how a better expression can be obtained as follows. First, factor
−
b
2
a
{\displaystyle -{\frac {b}{2a}}}
x
=
−
b
±
b
2
−
4
a
c
2
a
=
(
−
b
2
a
)
(
1
±
1
−
4
a
c
b
2
)
.
{\displaystyle {\begin{aligned}x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\&=\left(-{\frac {b}{2a}}\right)\left(1\pm {\sqrt {1-{\frac {4ac}{b^{2}}}}}\right).\end{aligned}}}
Q
2
=
a
c
b
2
.
{\displaystyle Q^{2}={\frac {ac}{b^{2}}}.}
Then
x
=
(
−
b
2
a
)
(
1
±
1
−
4
Q
2
)
{\displaystyle {\begin{aligned}x&=\left(-{\frac {b}{2a}}\right)\left(1\pm {\sqrt {1-4Q^{2}}}\right)\end{aligned}}}
Considering just the negative square root we have
x
1
=
(
−
b
2
a
)
(
1
−
1
−
4
Q
2
)
.
{\displaystyle {\begin{aligned}x_{1}&=\left(-{\frac {b}{2a}}\right)\left(1-{\sqrt {1-4Q^{2}}}\right).\end{aligned}}}
Multiplying the numerator and denominator by
1
+
1
−
4
Q
2
{\displaystyle 1+{\sqrt {1-4Q^{2}}}}
x
1
=
(
−
b
2
a
)
(
1
−
1
−
4
Q
2
)
(
1
+
1
−
4
Q
2
1
+
1
−
4
Q
2
)
=
(
−
b
2
a
)
(
1
−
1
+
4
Q
2
1
+
1
−
4
Q
2
)
=
−
2
b
Q
2
a
(
1
1
+
1
−
4
Q
2
)
=
−
2
b
(
a
c
b
2
)
a
(
1
1
+
1
−
4
Q
2
)
=
−
2
c
b
(
1
1
+
1
−
4
Q
2
)
=
−
c
b
(
1
1
2
+
1
2
1
−
4
Q
2
)
.
{\displaystyle {\begin{aligned}x_{1}&=\left(-{\frac {b}{2a}}\right)\left(1-{\sqrt {1-4Q^{2}}}\right)\left({\frac {1+{\sqrt {1-4Q^{2}}}}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=\left(-{\frac {b}{2a}}\right)\left({\frac {1-1+4Q^{2}}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2bQ^{2}}{a}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2b\left({\frac {ac}{b^{2}}}\right)}{a}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2c}{b}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {c}{b}}\left({\frac {1}{{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}}}\right).\end{aligned}}}
By defining
F
=
1
2
+
1
2
1
−
4
Q
2
{\displaystyle F={\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}}
we can write
x
1
=
−
c
b
F
.
{\displaystyle {\begin{aligned}x_{1}&=-{\frac {c}{bF}}.\end{aligned}}}
Note that as
Q
→
0
{\displaystyle Q\to 0}
F
→
1
{\displaystyle F\to 1}
Finding the Positive Root Using the Same Approach
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Turning now to the positive square root we have
x
2
=
(
−
b
2
a
)
(
1
+
1
−
4
Q
2
)
=
(
−
b
a
)
(
1
2
+
1
2
1
−
4
Q
2
)
=
−
b
F
a
.
{\displaystyle {\begin{aligned}x_{2}&=\left(-{\frac {b}{2a}}\right)\left(1+{\sqrt {1-4Q^{2}}}\right)\\&=\left(-{\frac {b}{a}}\right)\left({\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}\right)\\&=-{\frac {bF}{a}}.\end{aligned}}}
Using the two roots
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
y
=
a
x
2
+
b
x
+
c
=
a
(
x
2
+
b
a
+
c
a
)
=
a
(
x
−
x
1
)
(
x
−
x
2
)
=
a
(
x
+
c
b
F
)
(
x
+
b
F
a
)
.
{\displaystyle {\begin{aligned}y&=ax^{2}+bx+c\\&=a\left(x^{2}+{\frac {b}{a}}+{\frac {c}{a}}\right)\\&=a\left(x-x_{1}\right)\left(x-x_{2}\right)\\&=a\left(x+{\frac {c}{bF}}\right)\left(x+{\frac {bF}{a}}\right).\end{aligned}}}
Q
{\displaystyle Q}
edit
