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||Définition d'une variable || Algèbre || Arithmetique || Logique ||La raison basée sur les règles et modelés||
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YOU are better than YOU think. Show yourself  how: 

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Read  logic chapters 1 to 5  in online volume Three Skills for Algebra  for greater skills & confidence in  work 
and study.

Read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention. 

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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
 Amazing, Amusing, Amorous,  Delicious, Delightful, Edifying, Strengthening Elixir. 
It eases work & learning difficulties Makes the hard easier. Opens eyes. Leads to greater precision.
in reading and
writing

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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Caution: Site advice is approximately correct, for some circumstances, not all. That leaves room for thought

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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.


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For online automated help in senior high school maths & calculus, visit  quickmath.com  For Automatic Calculus and Algebra Help with derivatives, integrals, graphs, linear equations, matrix algebra, visit calc101.com  With  overlap, each site quickmath & calc101offers a different range of services, some free, some not, all based on webmathematica. Good luck.

Quadratics

Here are a few details, an overview, that may help provide a context for notes and explanations elsewhere.

Factoring by inspection uses the equation (x+A)(x+B) = x2+(A+B)x + AB. To get the completing the square equation x2+2Qx = (x+Q)2 - Q2  take A = B = Q and then subtract Q2 from both sides. Taking B= -A gives  (x+A)(x-A) = x2 - A2  or more generally,  (C+A)(C-A) =  C2 - A2  The latter equation provides a means to factor the difference of two squares.

Memory Aid for (x+A)(x+B) = x2+(A+B)x + AB
used in factoring by inspection

x
+
A

x2

Bx
Ax AB

x     +     B

For x, A and B all positive, the area of the large rectangle is (x+A)(x+B) or the sum of the areas of the small rectangle. This implies (x+A)(x+B) = x2+(A+B)x + AB. 

The condition that x, A and B all be positive can be removed if one uses the distributive law twice to obtain this result

(x+A)(x+B)  =  x(x+B) +  A(x+B) 

=  (xx+xB) +  (Ax+AB) 

=  (xx+Bx) +  (Ax+AB) 

=  x2+ Bx +Ax + AB

= x2+(B+A)x + AB

= x2+(A+B)x + AB.

 

Memory Aid for Completing the Square Identity
x2+2Qx = (x+Q)2 - Q2

x

+

Q

x2

Qx
Qx Q2

x     +    Q

(x+Q)2 = x2+2Qx + Q2.
 

Quadratic Formula and Related Material

By completing the square, each quadratic ax2+bx+c = a[(x-q)2 + h ] with q = -b/(2a) and h = (4ac-b2)/(4a2). The graph of y =  a[(x-q)2 + h ] has an axis of symmetry with equation x = q. Putting x = q gives y =  aq2+bq+c = a[(q-q)2 + h ] = ah.  

The point with coordinates [q, ah] = [q, aq2+bq+c] is the vertex of the quadratic. It is the lowest point on the quadratic if a> 0 and it is the highest point if a < 0.  If a> 0 the quadratic opens upward. If a < 0, the quadratic opens downward.  

If h < 0, then (x-q)2 + h = 0 when and only when  (x-q)2 = -h or 
x-q =±   __
Ö-h
  or   x = q ±   __
Ö-h

This gives the first way to solve a[(x-q)2 + h ] = 0 or
 ax2+bx+c = 0 when ax2+bx+c = a[(x-q)2 + h ]. The solutions are equidistant from the axis of symmetry, the line x = q.

If the discriminant b2-4ac > 0 then h < 0 and solutions of the quadratic  equation ax2+bx+c  = 0 are also given by

x =
-b±   ______
Öb2-4ac

2a

These two values are x-intercepts for the graph of y = ax2+bx+c. They are equidistant from its axis of symmetry = q.  Here q = -b/(2a).

Special Case: If the discriminant b2-4ac = 0 then h = 0 and  the quadratic ax2+bx+c  = 0  on the axis of symmetry and there is only one x-intercept, namely x = -b/(2a)

If you are given that or show that ax2+bx+c = a(x +s)(x+r)  then   x = -s and x = -r give one or two x-intercepts of y = ax2+bx+c, and  the axis of symmetry is at  x =  -½(r+s) = -b/(2a), halfway between the two intercepts. You may show that  show that ax2+bx+c = a(x +s)(x+r) with factoring by inspection (if it works) or via two steps: completing the square and using the difference of two squares.

Graphing Quadratics - parabolas

One way to sketch or graph the quadratics y = ax2+bx+c or y =a[(x-q)2 + h ] is to plot points on the curve y = ax2+bx+c at the x-intercept or intercepts,  if any, and for  x = q,  x = q ± 1/4, x = q ± 1/2,  x = q ± 1, x =q ± 2, etc, and then join these points by a smooth curve. Use fewer points if time is short. Here x = q = -b/(2a) is the equation of the  axis of symmetry for the curve y = ax2+bx+c. Hint: Calculate the coordinates of these points and then choose a unit lengths for the y and x axes. The unit lengths or scale on each axis may be different.

Remark: Completing the Square and the quadratic formulas works with complex numbers as well.  But when were are graphing quadratics y = ax2+bx+c with x restricted to real values, complex roots are not allowed. They are extraneous.   Here we have a situation where an equation ax2+bx+c = 0 may have  solutions outside the set where they are meaningful.  

Problems: 

A.    (i)  Use the quadratic formula to solve x2-3x-4= 1. 
       (ii) Find the value of y on the axis of symmetry of the quadratic y =  x2-3x-4.
        (iii) Use the results of (i) and (ii) to sketch the curve  y =  x2-3x-4.

B.   Find the intersection points of the quadratic y = x2 and the line y = 3x+4.    

C.    (i) Sketch the curve pq = 1 in the first quadrant of the  pq plane.  
      (ii) Give the definition of ln(x) for x > 1.
       (iii) Shade in the area under this curve pq = 1 that gives or defines ln(4). 

D.   (Step I) Complete the square for  x2-6x-8  and simplify the result.
      (Step II) Use the result of step I and the difference of two squares to factor  x2-6x-8  
      (Step III) Use the result of step II to solve x2-6x-8  = 0

 

 

 

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