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Deriving the Quadratic Formula
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| Theorem A: If a quadratic ax2+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. |
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 B: If a quadratic ax2+bx+c = a[(x-h)2
+ k ] for some real numbers h and k with k < 0
then quadratic is reducible and its has one or two real roots,
namely
x = h + (-k)½ and x = h - (-k)½ These formulas for roots yield a single root x = h when and only when k = 0. Proof: 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 = (-k)½ = sqrt(-k) . Hence -k = w2 .
The foregoing gives, implies or yields
where r = h + w and s = h - w provide 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 w = sqrt(-k) = sqrt(9) = 3, and so implies
5[(x-2)2 -9] = 5(x - 2 -3)(x-2+3) = 5(x-5)(x+1) = 5(x-r)(x-s)
where r = h + w = 2 + 3 and s = h - w = 2 - 3 = -1.
Equation Solving Route: If k < 0, then (x-h)2 + k = 0 when and only when (x-h)2 = -k or| x-h =± | __ Ö-k |
or | x = h ± | __ Ö-k |
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 solutions are equidistant from the axis of symmetry, the line x = h.
The completing the square 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 ± | __ Ö-k |
provides the quadratic formula. The factorization route gives the same formula and at the same time, shows how to express quadratic expressions ax2+bx+c as a product of factors and thus goes a little further.
| Theorem C: Each quadratic expression
ax2+bx+c = a[(x-h)2 + k] where
provided the x2-coefficient a is non-zero (to avoid division by zero). Proof: The proof here follows the path taken in the proof of Theorem B.
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We will continue the chain of reasoning started in the last proof.
| Observe in the last proof if - k = | b2 -4ac | > 0 |
|
4a2 |
then
| ax2+bx+c = a [ | (x+ |
b |
)2 | _ | b2 -4ac | ] |
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| = a [ | (x + |
b |
)2 | _ | { | sqrt(b2 -4ac) | }2 | ] |
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| = a difference C2 - R2 of two squares | ||||||||
| ax2+bx+c = a [ | x + |
b + sqrt(b2 -4ac) |
] | [ | x + |
b - sqrt(b2 -4ac) |
] |
| which can be factored as C2 - R2 = (C+R)(C-R) | |||||||
The first square in C2 - R2 is
| C2 =(x + |
b |
)2 |
and the second square is
| R2 = { | sqrt(b2 -4ac) | }2 |
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The factorization
| ax2+bx+c = a [ | x + |
b + sqrt(b2 -4ac) |
] | [ | x + |
b - sqrt(b2 -4ac) |
] |
says
ax2+bx+c = = a(x -r)(x-s)
where
| r = - |
b + sqrt(b2 -4ac) |
|
-b - sqrt(b2 -4ac) |
and
| s = - |
b - sqrt(b2 -4ac) |
|
-b + sqrt(b2 -4ac) |
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) |
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|
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.
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 ½(r+s) = - b/2a = h. Here completing the square would lead to the difference of two squares. The quadratic formula gives
| r = - |
b - sqrt(b2 -4ac) |
|
-b - sqrt(d) |
and
| s = - |
b - sqrt(b2 -4ac) |
|
-b + sqrt(d) |
Further
ax2+bx+c = a(x -r)(x-s)
The property ½(r+s) = - 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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