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
Memory Aid for Completing the Square Identity
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| x + Q |
x2 |
Qx |
| Qx | Q2 | |
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x + Q |
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| (x+Q)2 = x2+2Qx + Q2. | ||
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
This gives the first way to solve a[(x-q)2 + h ] = 0 or |
If the discriminant b2-4ac > 0 then h < 0 and solutions of the quadratic equation ax2+bx+c = 0 are also given by
| x = |
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.
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.
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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