Chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
Copied from 'Pig and Other Tales', Edited by Doug French and Charlie Stripp, published by
the Mathematical association in 1997, ISBN 0 906588 38 3
Chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
C R Holmes
Mathematical Gazette, 74, October 1990
At school, I was introduced to complex numbers by considering a parabola
drifting upwards:
Here it has
Here it has
Here it has
real
equal roots.
two
two real distinct roots.
no real roots.
We were eventually shown that all quadratics have two roots, and shown
how to find them. But I wanted to see them on the graph! It was explained to
me that you can't show them on the graph--+hat if the parabola doesn't cut the
axis then it's just hard luck-you can calculate the roots, but not show them on
the graph. This was most unsatisfying: I felt sure there should be some
relationship between the roots and the diagram.
Now, many years later, I have found a solution. I can now "see" the roots of
a quadratic equation, given its graph-and I can do this instantly and easily!
We can still use the idea of a parabola drifting upwards. But, this time,
imagine the parabola drags with it a reflection of itself, like this:
,
,,
,
I
\
I
t
The reflection is tied ; the "nose" of the original parabola, so that, as the
original parabola rises, so does the reflection.
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Chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
We can use the reflection to "see" the roots, when the original parabola has
drifted above the x-axis.
(i)
l=i-6x+5
has solutions3
+
2.
(ii)
!=i
hassolutions3
t
0.
-6x+9
(iii) t=*-6x+13
has solutions3
!
2i.
In general by imagining a reflection of the parabola in qit way one can see
its imiginary roots: the interested reader can easily check this.
C R HOLMES
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Chapter 6b: THE COMPLEX ROOT OF A QUADRATIC FROM ITS GRAPH
Chapter 6b: THE COMPLEX ROOT OF A QUADRATIC
FROM ITS GRAPH
S F G Wessels
Mathematical Gazette, 65, March 1981
Letthecomplexrootsof thequadraticequationf (x) =
beb + ic.Then
f(x)= (x - b + ic)(x - b - ic\
=x2 - 2bx + b2 + c2
*
+ px
*q=O
-
b)2 + c2.
Therefore b is the x value where the parabola y = f (x) has a minimum value.
Draw the graph ) = f (x). Then (b, C) is ttre minimum point of the curve.
Theline = 2c2 cutsy=f(r)inthepoints C(b c,2c2)andD(b + c,
2cz) sine,
=(x
-
!
f(x)= (x-bf +d=k2
(x-b)2=f
x=b+corx=b-c.
From the graph we can find c2 and we can thus draw the line
y = 2c2.
c+c-1
From the graph we can now see the values of b and c.
S F G WESSELS
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ROOTS OF A QUADRATIC
Questions on chapter 6a: IMAGINE THE
1. (a)
(b)
2.
4.
How does this form provide the roots?
If an equation
x=
3.
By factorising, confirm the solutions in cases (i) and (ii).
has real roots, explain why the solutions
m ! n' where
T" of the form
m'is the line of symmetry of the corresponding graph'
By using the quadratic formula, show case (iii) has no real solutions'
(a) Complete the square in case (iii).
(b) How can you deduce from this form that there are no real roots?
(c) How can the complex roots be identified from this form?
6x + 13 to explainwhy.the t:{}"tt*curve in
these
case (iii) has an equation which can be written ! = 4 - (x - 3)'. Use
squarc forrns for the equations of the two curves to explain the reflection
"o*pfLt"a
method for finding imaginary roots.
5. Use the compteted square form of x2
-
for a quadratic can be written y = F2 + bx + c' Rearrange this by
comileting the square. Rewrite to give an expression for x when ;l = 0' What
6. The general form
condlitions must D and c
Questions on ChaPter 6b:
I
fulfil for there to be real roots?
THE COMPLEX ROOT OF A QUADRATIC
FROM ITS GRAPH
equation f (x) = (x ' b)2 + c2 given i1 this article,
7. By reference to the general
-y
eiplain why the line = 8 aliows u$ to find the complex roots of the equation
x2 - 6x + 13 = 0 in question 6.
g. Use a similar method to the one illustrated in the article to find the complex roots of:
(a) x2+6x+25=0
(b) 2x2+3x*2=0.
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Answers to chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
Answers to Chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
1. (a) Case(i) x2 - 6x+ 5 = (x Case (ii) x2 - 6x + 9 = (x -
5)(x - l)thereforerootsat x = Sandx = l"
3)(x - 3) therefore double root at x = 3.
(b) By equating each bracket to z/Jro, as their product must equal zero where
roots
occur.
2.
This uses the symmerical properties of the parabola'
The roots are symmetrically placed on either side of x = nt.
(Also consider how completing the square or using the quadratic 'formula' works.)
3.
* + bx + c = Ohasnorealroots if bz - 4ac < O'
lni - 6x+ 13 = 0,a=l,b=-6 and c=13, sob2-4ac=-16<0' Therefore the
equation has no real roots.
4.
(a) y = f - Ox + 13 = Y = (x - 3)2 + 4.
(b) By completing the square it can be seen that the minimum value of y is 4. This tells
13 = 0
us that the curve never crosses the x-axis and so the equation x2 - 6x +
has no real roots.
(c) (x
5.
- 3f * 4 =0 +
x
- J - 64 = +2i + x = 3!2i'
*-Ox+13= (x-3)2+4.
when the curye
is reflected in the line ! = 4' the equation
The roots corresponding
!= 4+(x-3)2 to !=4-(*:3)'.
3 + 2i and3 ! 2'
(x
0arethen
3f
+ (x - 3)2 = Oand 4 - =
-4
6. !=x2+bx+c.
Completingthesquare,Y
= (,
.if
changes from
to the equations
+c-- b24
)=o=(,.if=*Z-,.
Rearranging gives
, = -#,
so b2
-
4c
)
0 for there to be real roots'
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FROM ITS GRAPH
Answers to chaprer 6b: THE COMPLEX ROOT OF A QUADRATIC
Answers to chapter 6b: THE COMPLEX ROOT OF A QUADRATIC
-
7.
FROM ITS GRAPH
y = (x - 3\2 + 4'
l=i-6x+l3iswrittenasacompletedsquareitbecomes
2cz-becomes the line ,l = 8'
The + 4 in this case represents the +c2, so the line ! =
When
which
inY
4.
- "*,
UV
"o*parini
with the other method, be thought of as reflection of the x-axis
g. (a) y =i-+6x+25, complete the squarei y=(x+3)2 +16, ry:PY.i:.::
G . ,f'Jo = 0 ;;ymmetrical'auoy- x = -3. *g with
ryt-f
'* 9T11Yl
16 to give c = !4,
(-3,16). The roots of ,t equation are found by equating c2
"
so giving x = -3 + 4i andx = -3 - 4i'
(b) y J- ui * 3! + ?, divide through by 2: =. \trx +
* = <; * if_+ rt, ,o rhe roots iv-r*tri$ a!11 1
i t
,
r':?T!',"t-i:h:,:1Y,:
-"
Ttl:1.:
: -,t'_i1.:h.:
c2 with rt to give
found by equating
*"
ar"
;i
fi;;;
i+','iui.
i;,X
* - i4.
l = =*,"or-'= * + i*orx = "quution
i
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