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6. Quadratics

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.x = 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

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.

Algebra, Odds & Ends,

Odd and Ends, Essays

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