Transcript Polynomials A polynomial is an algebraic expression made up of different powers of a particular letter. The degree of the polynomial is the highest.

Polynomials
A polynomial is an algebraic expression made up of
different powers of a particular letter.
The degree of the polynomial is the
highest power in the expression
Examples
3x4 – 5x3 + 6x2 – 7x - 4
Polynomial in x of degree 4
7m8 – 5m5 – 9m2 + 2
Polynomial in m of degree 8
w13 – 6
Polynomial in w of degree 13
Evaluating Polynomials
Suppose that g(x) = 2x3 - 4x2 + 5x - 9
Substitution Method
g(2) = (2 × 2 × 2 × 2) – (4 × 2 × 2 ) + (5 × 2) - 9
= 16 – 16 + 10 – 9
= 1
this requires 9 calculations
Nested or Synthetic Method
This involves using the coefficients and requires fewer
calculations so is more efficient.
It can also be carried out quite easily using a calculator.
g(x) = 2x3 – 4x2 + 5x – 9
2, -4,
Coefficients are
g(2) =
2
2
5, -9
-4
5
-9
4
0
5
10
1
0
Factor Theorem
If
(x – a) is a factor of the polynomial f(x)
then
f(a) = 0 and a is a root of y = f(x)
Consider
f(x) = x2 – 3x – 18
Factorise
f(x) = x2 – 3x – 18
f(x) = (x – 6)(x + 3)
f(6) = 62 – 3 × 6 – 18
f(6) = 0
f(– 3) = (–3)2 – 3×(–3 ) – 18
f(– 3) = 0
Factor Theorem
Try f(5) → 5
Consider f(x) = x3 – 6x2 – x + 30
1
-6
-1
30
5
-5
-30
×5
×
5
×
5
1
-1
-6
0
These numbers give the
coefficients of the other factor
Other factor is x2 – x - 6
So x3 – 6x2 – x + 30 = (x – 5)(x2 – x – 6)
= (x – 5)(x – 6)(x + 1)
f(5) = 0 so
(x – 5)
is
a factor
Example
x3 + 3x2 – 10x - 24
Factorise
We need some trial & error with factors of –24
ie
+/-1, +/-2, +/-3 etc
f(2)
1
3
2
×2
×2
5
1
-10
10
0
×2
-24
0
– 24
f(2) is not equal to zero
Now try f(-2)
If 1 or –1 is a
root the
coefficients will
add to 0
Example
f(-2)
Factorise
1
x3 + 3x2 – 10x - 24
3
-10
-2
-2
×-2 ×-2
×-2
1
1
-12
-24
24
0
f(-2) = 0
(x + 2) is a factor
x3 + 3x2 – 10x - 24 = (x + 2)(x2 + x – 12)
x3 + 3x2 – 10x - 24 = (x + 2)(x + 4)(x – 3)
The roots or zeros of a polynomial tell us
where it cuts the x-axis, ie where f(x) = 0
So the graph of y = x3 + 3x2 – 10x – 24
cuts the x-axis at –2, –4 and 3
Factorising Higher Orders
Solve
f(-1)
x4 + 2x3 – 8x2 – 18x – 9Coefficients
=0
don’t add up to 0
but
of
1
2
-8
-18combinations
-9
f(-1)
=0
the
numbers
do,
9
-1
9
-1
so
(x
+
1)
so
try
f(-1)
×– 1 ×– 1
×– 1 ×– 1
a factor
-9
1
1
-9
0
x4 + 2x3 – 8x2 – 18x – 9 = (x + 1)(x3 + x2 – 9x – 9)
Now repeat the process for
x3 + x2 – 9x – 9
x4 + 2x3 – 8x2 – 18x – 9 = (x + 1)(x3 + x2 – 9x – 9)
Now repeat the process for
x3 + x2 – 9x – 9
f(3)
1
1
1
3
×3
4
-9
12
×3
3
-9
9
×3
0
f(3) = 0
so (x – 3)
is a factor
x4 + 2x3 – 8x2 – 18x – 9 = (x + 1)(x – 3)(x2 + 4x + 3)
= (x + 1)(x – 3)(x + 1)(x + 3)
So roots are 3, –1 (twice), –3
Linear Factors in the form (ax + b)
If
(ax + b) is a factor of the polynomial f(x)
then f( -b /a ) = 0
Suppose f(x) = (ax + b)(………..)
If f(x) = 0 then (ax + b)(………..) = 0
So (ax + b) = 0 or (…….) = 0
so ax = -b
so x = -b/a
NB: When using such factors we need to take care
with the other coefficients
Linear Factors in the form (ax + b)
Example
Show that (3x + 1) is a factor of g(x) = 3x3 + 4x2 – 59x – 20
and hence factorise the polynomial completely
Since (3x + 1) is a factor then g(-1/3) should equal zero
g(-1/3)
3
4
-59
-20
3
-1
3
-1
-60
20
0
The other factor is
3x2 + 3x – 60
g(- 1/3) = 0
so (x + 1/3)
is a factor
g(x) = 3x3 + 4x2 – 59x – 20
= (x + 1/3) (3x2 + 3x – 60)
= (x + 1/3)3×(x2 + x – 20)
= (3x + 1)(x2 + x – 20)
= (3x + 1)(x + 4)( x – 5)
Roots are –1/3, – 4 and 5
Missing Coefficients
Given that (x + 4) is a factor of the polynomial
f(x) = 2x3 + x2 + kx – 16
find the value of k
and hence factorise f(x)
Since (x + 4) a factor
f(-4)
2
2
1
-8
-7
k
28
(k + 28)
then
f(-4) = 0 .
-16
(-4k – 112)
(-4k – 128) = 0
4k + 128 = 0
4k = –128
k = –32
f(x) = 2x3 + x2 + kx – 16
k = –32
f(x) = 2x3 + x2 – 32x – 16
f(-4)
2
2
1
-8
-7
–32
k
28
(k + –4
28)
f(x) = 2x3 + x2 – 32x – 16
f(x) = (x + 4)(2x2 – 7x – 4)
f(x) = (x + 4)(2x + 1)( x – 4)
-16
(-4k – 112)
(-4k – 128) = 0
Missing Coefficients
(x – 4) is a factor of f(x) = x3 + ax2 + bx – 48 while
f(-2) = -12. Find a and b and factorise f(x) completely.
(x – 4) a factor so f(4) = 0
f(4)
1
1
a
4
b
(4a + 16)
-48
(16a + 4b + 64)
(a + 4) (4a + b + 16) (16a + 4b + 16) = 0
16a + 4b + 16 = 0 (4)
4a + b + 4 = 0
4a + b = -4
Missing Coefficients cont’d
f(-2) = -12 so
f(-2)
1
1
a
-2
b
(-2a + 4)
(a - 2) (-2a + b + 4)
-48
(4a - 2b - 8)
(4a - 2b - 56) = -12
4a - 2b - 56 = -12(2)
2a - b - 28 = -6
2a - b = 22
We now use simultaneous equations ….
add
4a + b = – 4
2a – b = 22
6a = 18
a=3
Using 4a + b = – 4
12 + b = – 4
b = – 16
When (x – 4) is a factor the quadratic factor is
x2 + (a + 4)x + (4a + b + 16) = x2 + 7x + 12
= (x + 4)(x + 3)
So f(x) = (x – 4)(x + 3)(x + 4)
Finding a Polynomial from its Roots or Zeros
Consider f(x) = x2 + 4x – 12
Roots  f(x) = 0
x2 + 4x – 12 = 0
(x + 6)(x – 2) = 0
Roots
x = -6 or x = 2
yy
–8–8
–8
–6–6
–6
–4
–4
–2
xx
2
–2
2
4
4
–10
–10
–20
–20
–30
–40
–40
The roots are the
intersection of the graph
with the x-axis
As can be seen, all scaled (or multiple) versions
of the function pass through the same roots
y = kf(x) has the same roots for all values of k
Finding a Polynomial From Its Zeros
If a polynomial f(x) has roots/zeros at a, b and c
it has factors (x – a), (x – b) and (x – c)
It can be written as
f(x) = k(x – a)(x – b)(x – c)
y
y = f(x)
30
-2
1
5
f(x) = 3(x + 2)(x – 1)(x – 5)
Roots are –2 , 1 and 5  f(x) = k(x + 2)(x – 1)(x – 5)
Also f(0) = 30
 30 = k( 2)(– 1)(– 5)
 30 = k × 10
k=3
x
Roots
(0, )
Max. Point
(0, )
f(x) = x2 + 4x + 3
f(-2) =(-2)2 + 4x(-2) + 3
= -1
a>0
x=
Mini. Point
Line of Symmetry
half way
between roots
a<0
x=
Line of Symmetry
half way
between roots
Evaluating
Graphs
Quadratic Functions
y = ax2 + bx + cc
Decimal
places
Cannot Factorise
Factorisation
ax2 + bx + c = 0
e.g. (x+1)(x-2)=0
Roots
x = -1 and x = 2
b  (b2  4ac)
x
2a
Roots
x = -1∙2 and x = 0∙7
Completing the Square
This is a method for changing the format
(the way we write)
of a quadratic equation
In this form we can easily sketch or
read off key information
Completing the square format looks like
f(x) = a(x + b)2 + c
Completing the square reverses FOIL
x2 + 2ax + a2
→
(x + a)2
Isolate the x and
x2x+2 6x
is part
(x + 3)2
[ x2 + 6x] + 1
terms
in of
square
brackets
Still
equal to
2
2
= [(x + 3) – 9 ] + 1
x + 26x + 9
x + 6x
= (x +
3)2
–8
y
x
y = (x + 3)2 – 8
A ‘HAPPY’ parabola with minimum
turning point x = –3 , y = –8
(–3, –8)
→
x2 + 2ax + a2
[ x2
=
= (x –
the
x2 – Isolate
2x is part
of x
(xand
– 1)2
– 2x ] – 7
[ (x – 1)2 – 1 ]
1)2
(x + a)2
–7
x2 terms in square
x2brackets
– 2x
+ 1 to
Still
equal
x2 – 2x
–8
y
x
y = (x – 1)2 – 8
A ‘HAPPY’ parabola with minimum
turning point x = 1 , y = –8
(1, –8)
[ 3x2
– 12x ] + 5
= 3[x2 – 4x]
+5
= 3[(x – 2)2 – 4 ] + 5
Take out common factor 3
x2 – 4x is part of (x – 2)2
x2 – 4x + 4
= 3(x – 2)2 – 12 + 5
y
= 3(x – 2)2 – 7
x
y = 3(x – 2)2 – 7
A ‘HAPPY’ parabola with minimum
turning point x = 2 , y = –7
(2, –7)
1[ + 10x – x2
]
= –1[x2 – 10x] + 1
= –1[(x – 5)2 – 25 ] + 1
Take out common factor –1
x2 – 10x is part of (x – 5)2
x2 – 10x + 25
= –(x – 5)2 + 25 + 1
= –(x – 5)2 + 26
(5, 26)
y = –(x – 5)2 + 26
x
A ‘SAD’ parabola with maximum
turning point x = 5 , y = 26
y
1
Consider the fraction
n
1
2
1
7
Is bigger
than
1
10
Is smaller
than
1
3
Is bigger
than
Is smaller
than
1
50
7
1
 1  7
1
1
7
1
Consider the fraction
f ( x)
The maximum value of the fraction will
occur when f(x) is a minimum
The minimum value of the fraction will
occur when f(x) is a maximum
Express x2 + 6x + 13 in the form (x + p)2 + r.
Hence state the maximum value of
1 of 2
1
f ( x)  2
x  6 x  13
[ x2
+ 6x ] + 13
= [(x + 3)2 – 9
= (x +
3)2
]+
x2 + 6x is part of (x + 3)2
13
x2 + 6x + 9
y
+4
x
y = (x + 3)2 + 4
A ‘HAPPY’ parabola with minimum
turning point x = –3 , y = 4
(–3, 4)
Express x2 + 6x + 13 in the form (x + p)2 + r.
Hence state the maximum value of
2 of 2
1
f ( x)  2
x  6 x  13
y
y = (x + 3)2 + 4
x
The maximum value of a
fraction occurs when the
denominator is as small as
possible, ie a minimum.
(–3, 4)
The minimum value of x2 + 6x + 13 is 4
So MAX f(x) is 1/4
Express 3 – 8x – 4x2 in the form p(x + q)2 + r.
Hence state the minimum value of
10
f ( x) 
3  8 x  4 x2
3[ – 8x – 4x2
= –4[x2 + 2x]
Take out common factor –4
]
x2 + 2x is part of (x + 1)2
+3
= –4[(x + 1)2 – 1 ] + 3
= –4(x +
1)2
x2 + 2x + 1
(–1, 7)
+4+3
= –4(x + 1)2 + 7
 MAX value is 7
x
So
MIN
value of f(x) is 10/7
y
Complete the square for x2 + 2x + 3 and hence
sketch function and show it is never negative.
So this quadratic has no roots
y
and
is
always
positive.
2
= (x + 1) + 22
Insert [ ]yaround
[ x2 + 2x]+ 3
Compare with (a + b)
variables
= [(x +
1)2
– 1] + 3
= (x + 1)2 – 1 + 3
= (x + 1)2 + 2
x2 + 2x is created
2 + (0
, +3)1
→
x
2x
2
from (x + 1)
Keep expression in
(–1– ,12)
[ ] equal by
A ‘HAPPY’ parabola with a
MINIMUM at (– 1 , 2)
x
Complete the square for 2x2 – 8x + 9, hence sketch
the function and show that it is never negative.
So this quadratic has no roots
and is always
positive.
y
[ ] round variables
[2x2 – 8x]+ 9
y = 2(x - 2)2 + 1
Take out common factor
= 2[x2 – 4x] + 9
x2 - 4x → (x - 2)2
= 2[(x – 2)2 – 4] + 9
(0 , 9)
= 2(x – 2)2 – 8 + 9
= 2(x – 2)2 + 1
– 4 to equalise [ ]
out [ ]
(2Multiply
, 1)
A ‘HAPPY’ parabola with a
MINIMUM at (2 , 1)
x
Complete the square for 7 + 6x – x2
and hence
function.
Noticesketch
that this
quadratic has
[–
x2
+ 6x]+ 7
two roots andy has positive
(3 , 16)and
] round variables
negative [portions.
= –1[x2 – 6x] + 7
Take out common factor – 1
(0 , 7)
= –1[(x – 3)2 – 9] + 7
x2 – 6x → (x – 3)2
Multiply out [ ]
= –1(x – 3)2 + 9 + 7
= –1(x –
3)2
+ 16
A ‘SAD’ parabola with a
MAXIMUM at (3 , 16)
x
Using Discriminants
Given the general form for a quadratic function.
f(x) = ax2 + bx + c
We can calculate the value of the DISCRIMINANT
b2 – 4ac
This gives us valuable information
about the roots of the quadratic function
The Discriminant
Given
ax2 + bx + c = 0
 b  b  4ac
x
2a
2
b2 – 4ac
is the part under the square root sign
The discriminant
The discriminant → Roots of a quadratic Function
There are 3 possible scenarios
 b  b  4ac
x
2a
2
2 distinct
real roots
1 real root
‘equal roots’
No real roots
discriminant
discriminant
discriminant
(b2- 4ac > 0)
(b2- 4ac = 0)
(b2- 4ac < 0)
The Discriminant and Rational or Irrational roots
Given ax2 + bx + c = 0
 b  b  4ac
x
2a
2
Roots can only be
rational
if discriminant
b2 – 4ac
is a perfect square
Rational  a number which can be written as a fraction
Irrational  a number which, as a decimal,
never ends or repeats.
The discriminant of a quadratic equation is 23.
Here are two statements about this quadratic equation:
(1) the roots are real;
(2) the roots are rational.
 b  b 2  4ac
x
2a
Discriminant, b2 – 4ac = 23
b2 – 4ac = > 0  real roots
23 is not a perfect square
 roots are irrational
Both statements
are true
For what value of k does the equation
x2 – 5x + (k + 6) = 0 have equal roots?
a = 1, b = – 5,
c=k+6
b2 – 4ac = (– 5)2 – 4 × 1 × k + 6
= 25 – 4k – 24
= 1 – 4k

Equal roots 
1 – 4k = 0

4k = 1

k = 1/4
b2 – 4ac = 0
Show that the roots of
(k - 2)x2 – (3k - 2)x + 2k = 0 are always real
a = (k – 2)
b = – (3k – 2)
c = 2k
b2 – 4ac = [– (3k – 2) ]2 – 4(k – 2)(2k)
= 9k2 – 12k + 4 –Can
8k2 +never
16k be negative
no matter what value
= k2 + 4k + 4
is chosen for k
= (k + 2)2
Since square term b2 – 4ac ≥ 0 and roots ALWAYS real.
Show that the equation (1 – 2k)x2 – 5kx – 2k = 0 has
real roots for all integer values of k.
Real roots  b2 – 4ac ≥ 0
a = (1 – 2k), b = – 5k, c = – 2k
b2 – 4ac = (– 5k)2 – 4(1 – 2k) × – 2k
An Integer
is any member
of the set
2
2
2 + 8k
=…….
25k -3
+ 8k
–
16k
→
9k
, -2 , -1 , 0 , 1 , 2 , 3 ……. 8
– /9
2
Consider graph of y = 9k + 8k = k(9k + 8)
A “HAPPY” parabola through 0 and – 8/9
Hence equation has real roots for all integer k
Using the Discriminant
Find the value p given that 2x2 + 4x + p = 0 has real roots
Real roots  b2 – 4ac ≥ 0
a=2
16 – 8p ≥ 0
– 8p ≥ – 16
b=4
c=p
÷ by negative
reverses
inequality
p≤2
The equation has real roots when p ≤ 2.
Find w given that x2 + (w – 3)x + w = 0 has non-real roots
For non-real roots 
a = 1 b = (w – 3) c = w
(w – 3)2 – 4w < 0
w2 – 6w + 9 – 4w < 0
w2 – 10w + 9 < 0
b2 – 4ac < 0
Sketch graph of
y = (w – 9)(w – 1).
9
A1“HAPPY” parabola
passing through – 1
and
the
y =9w2on
– 10w
+ 9 x-axis
(w – 9)(w – 1) < 0
From graph non-real roots occur when 1 < w < 9
Tangency → a line cuts a curve in one point only
Tangency  equal roots
b2 – 4ac = 0
Line cuts curve in
two distinct points
tangency
Line cuts curve in
only one point
Line does not cut
curve
Prove that the line y = x + 1 is a tangent to
the curve y = x2 + 3x + 2.
Curves will intersect where y = y, ie solve equations
x2 + 3x + 2 = x + 1
x2 + 3x + 2 – x – 1 = 0
x2 + 2x + 1 = 0
b2 – 4ac = (2)2 – 4(1)(1)
= 0 (equal roots)
→ (x + 1)(x + 1) = 0
→ x = – 1 (twice)
Equal roots
Since only 1 real root line is tangent to curve.
Prove that y = 2x – 1 is a tangent to the curve y = x2
and find the intersection point
x2 = 2x - 1
x2 – 2x + 1 = 0
(x – 1)2 = 0
x = 1, twice
at point of intersection
b2Only
- 4ac1 root
2 – 4(1)(1)
hence
tangent
= (-2)
= 0 hence tangent
When x = 1 then y = (1)2 = 1 so intersection point is (1,1)
Or x = 1 then y = 2×1 - 1 = 1so intersection point is (1,1)
Find the equation of the tangent to y = x2 + 1
that has gradient 2.
The tangent must have the form y = 2x + k
Tangent and curve meet
where y = y
x2 + 1 = 2x + k
x2
– 2x + (1 – k) = 0
a=1
Equal roots  b2 – 4ac = 0
b=–2
c = (1 – k)
b2 – 4ac = (– 2)2 – 4(1 – k)
So 4 – 4 + 4k = 0
k=0
Tangent equation is y = 2x
Show that the line with equation y = 2x + 1 does not
intersect the parabola with equation y = x2 + 3x + 4
Solve equations to find point of intersection
x2 + 3x + 4 = 2x + 1
x2 + x + 3 = 0
Calculate discriminant
a = 1, b = 1, c = 3
b2 – 4ac = 12 – 4×1×3
= – 11
<0
No real roots
Line does not intersect curve
The diagram shows a sketch of a
parabola passing through (–1, 0),
(0, p) and (p, 0).
a) Show that the equation of the
parabola is y = p + (p – 1)x – x2
y = k(x + 1)(x – p)
Use point (0, p) to find k
p = k(0 + 1)(0 – p)
 p = – kp  k = – 1
y = – 1(x + 1)(x – p)
y = – 1(x2 + x – px – p)
y=–
x2 –
x + px + p
y = p + (p – 1)x – x2
The diagram shows a sketch of a
parabola passing through (–1, 0),
(0, p) and (p, 0).
a) Show that y = p +(p – 1)x – x2
b) For what value of p will the line
y = x + p be a tangent to this curve?
p +(p – 1)x – x2 = x + p
b2 – 4ac = 0
for tangency
(p – 2)x – x2 = 0
b2 – 4ac = (p – 2)2 – 4×–1× 0
b2 – 4ac = (p – 2)2
p=2
If x2 – 8x + 7 is written in the form (x – p)2 + q,
what is the value of q?
[ x2 –
8x ]+ 7
= [ (x – 4)2 – 16 ] + 7
(x – 4)2 = x2 – 8x + 16
= (x – 4)2 – 16 + 7
= (x – 4)2 – 9
q= – 9
A function f is defined on the set of real numbers by
f(x) = x3 – x2 + x + 3.
What is the remainder when f(x) is divided by (x – 1)?
1
1
1
–1
1
3
1
0
1
0
1
4
Remainder = 4
The discriminant of a quadratic equation is 23.
Here are two statements about this quadratic equation:
(1) the roots are real;
(2) the roots are rational.
 b  b 2  4ac
x
2a
Discriminant, b2 – 4ac = 23
b2 – 4ac = > 0  real roots
23 is not a perfect square
 roots are irrational
Both statements
are true
If f(x) = (x – 3)(x + 5), for what values of x is the graph of
y = f(x) above the x-axis?
Sketch parabola, y = (x – 3)(x + 5)
Cuts x-axis at y = 0
 x = 3 and x – 5
A ‘HAPPY’ parabola
x
Graph above x-axis
when
x < –5 and x > 3
–5
3
Functions f, g and h are defined on the set of real numbers by
• f(x) = x3 – 1
• g(x) = 3x + 1
• h(x) = 4x – 5.
(a) Find g( f(x)).
(b) Show that g( f(x)) + xh(x) = 3x3 + 4x2 – 5x – 2.
(c) (i) Show that (x – 1) is a factor of 3x3 + 4x2 – 5x – 2.
(ii) Factorise 3x3 + 4x2 – 5x – 2 fully.
(d) Hence solve g( f(x)) + xh(x) = 0.
g(f(x)) = g(x3 – 1) = 3(x3 – 1) + 1 = 3x3 – 2
xh(x) = x(4x – 5) = 4x2 – 5x
g( f(x)) + xh(x)
= 3x3 + 4x2 – 5x – 2
(i) Show that (x – 1) is a factor of 3x3 + 4x2 – 5x – 2.
(ii) Factorise 3x3 + 4x2 – 5x – 2 fully.
(c)
(d) Hence solve g( f(x)) + xh(x) = 0.
1
3
3
4
–5
–2
3
7
2
7
2
0
So 1 is a root,
(x – 1) a factor
3x3 + 4x2 – 5x – 2 = (x – 1)(3x2 + 7x + 2)
= (x – 1)(x + 2)(3x + 1)
g( f(x)) + xh(x) = 0
 x = 1, –2 or – 1/3
Show that y = mx meets the parabola y = x2 + 1 at points given
by x2 – mx + 1 = 0.
Find values of m so that the line is a tangent and write down
the equations of these lines.
x2 + 1 = mx
x2
– mx + 1 = 0
a=1
b=–m
Curves meet where y = y
Tangency, Equal roots
 b2 – 4ac = 0
c=1
b2 – 4ac = (– m)2 – 4  1  1
So m2 – 4 = 0
m2 = 4
m = –2 or 2
Tangent equations
y = – 2x or y = 2x
For what value of t is the line x – y + t = 0 a tangent to y2 = 4x
x–y+t=0
x+t=y
Curves meet where y = y
(x + t )2 = 4x
 x2 + 2xt + t 2 – 4x = 0
Tangency, Equal roots
 b2 – 4ac = 0
 x2 + x(2t – 4 ) + t 2 = 0
b = 2t – 4
a=1
c=t2
b2 – 4ac = (2t – 4)2 – 4  1  t 2
4t 2– 16t + 16 – 4t 2 = 0
– 16t + 16 = 0
t=1
Find k if y = x – k is a tangent to x2 + y2 = 8
Find the point of contact
x2 + (x – k)2 = 8
Curves meet where y = y
 x2 + (x2 – 2kx + k 2) – 8 = 0
 2x2 – 2kx + k 2 – 8 = 0
a = 2 b = –2k c = k 2 – 8
Tangency, Equal roots
 b2 – 4ac = 0
b2 – 4ac = ( –2k)2 – 4  2  ( k 2 – 8 )
4k 2 – 8k2 + 64 = 0
–4k2 + 64 = 0 ÷ – 4
k2 – 16 = 0  k = –4 or 4
Find k if y = x – k is a tangent to x2 + y2 = 8
Find the point of contact
x2 + (x2 – k)2 = 8
 x2 + (x2 – 2kx + k 2) – 8 = 0
 2x2 – 2kx + k 2 – 8 = 0
When k = – 4 (1) becomes
 k = –4 or 4
for tangency
(1)
Repeating for k = 4 gives
2x2 + 8x + 8 = 0
2(x2 + 4x +4) = 0
2(x + 2)(x + 2) = 0
x = – 2 twice
y = – 2 – (-4) = 2
x = 2 twice
(–2 , 2)
y = 2 – 4 = –2 (–2 , –2)