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YOU are better than YOU think. Show yourself how:
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-/[]\- Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite. Logic
Mastery Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck. After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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-/[]\- What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts. Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice. |
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 | ] |
|
|
| = 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 |
|
|
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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Vec & Cmplx No Applet
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