SOLUTION: Could somebody guide me about this which is about the relation between the roots and the coefficients? For what values of m will the equation {{{x^2-4x+7=m(x-1)}}} have a) one ro

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Could somebody guide me about this which is about the relation between the roots and the coefficients? For what values of m will the equation {{{x^2-4x+7=m(x-1)}}} have a) one ro      Log On


   

Question 202022: Could somebody guide me about this which is about the relation between the roots and the coefficients?
For what values of m will the equation x%5E2-4x%2B7=m%28x-1%29 have
a) one root the reciprocal of the other,
b) one root equal zero
c) equal roots
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Could somebody guide me about this which is about the relation between the roots and the coefficients?
For what values of m will the equation x%5E2-4x%2B7=m%28x-1%29 have
a) one root the reciprocal of the other,
b) one root equal zero
c) equal roots

To do any of those we have to first get

x%5E2-4x%2B7=m%28x-1%29 in the form x%5E2%2BAx%2BB=0
x%5E2-4x%2B7=mx-m
x%5E2-4x-mx%2B7%2Bm=0
x%5E2-%284%2Bm%29x%2B%287%2Bm%29=0

----------------

a) one root the reciprocal of the other

Suppose one root is r and the other is 1%2Fr, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

%28x-r%29%28x-1%2Fr%29=0

x%5E2-%28r%2B1%2Fr%29x%2B1=0

Therefore this must be the same equation as 

x%5E2-%284%2Bm%29x%2B%287%2Bm%29=0

So we equate like parts:

system%28-%28r%2B1%2Fr%29=-%284%2Bm%29%2C1=%287%2Bm%29%29

or simplifying:

system%28r%2B1%2Fr=4%2Bm%2C-6=m%29

To check that, we substitute -6 
for m:

x%5E2-4x%2B7=m%28x-1%29
x%5E2-4x%2B7=-6%28x-1%29
x%5E2-4x%2B7=-6x%2B6
x%5E2%2B2x%2B1=0
%28x%2B1%29%28x%2B1%29=0
x%2B1=0, x%2B1=0    
x=-1  , x=-1

So both roots are equal and -1, but -1 is the
reciprocal of -1.

So the answer to (a) is m=-6
-------------

b) one root equal zero

Suppose one root is 0 and the other is r, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

%28x-0%29%28x-r%29=0
x%28x-r%29=0
x%5E2-rx=0

Therefore this must be the same equation as 

x%5E2-%284%2Bm%29x%2B%287%2Bm%29=0

So we equate like parts:

system%28-r=-%284%2Bm%29%2C0=%287%2Bm%29%29

Simplifying:

system%28r=4%2Bm%2C-7=m%29

So we see that m must be -7.

To check that, we substitute -7 
for m:

x%5E2-4x%2B7=m%28x-1%29
x%5E2-4x%2B7=-7%28x-1%29
x%5E2-4x%2B7=-7x%2B7
x%5E2%2B3x=0
x%28x%2B3%29=0
x=0, x%2B3=0    
           x=-3

One root is 0, so we are correct.

So the answer to (b) is m=-7

-------------

c) equal roots

Suppose one root is r and the other is also r, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

%28x-r%29%28x-r%29=0
%28x-r%29%28x-r%29=0
x%5E2-2rx%2Br%5E2=0

Therefore this must be the same equation as 

x%5E2-%284%2Bm%29x%2B%287%2Bm%29=0

So we equate like parts:

system%28-2r=-%284%2Bm%29%2Cr%5E2=%287%2Bm%29%29

Simplifying:

system%282r=4%2Bm%2Cr%5E2=7%2Bm%29

Solve both equations for m

system%28m=2r-4%2Cm=r%5E2-7%29

Equate the right sides since both equal m:

2r-4=r%5E2-7
-r%5E2%2B2r%2B3=0
r%5E2-2r-3=0
%28r%2B1%29%28r-3%29=0
r%2B1=0, r-3=0    
r=-1 , r=3

Substituting r=-1 into

m=2r-4
m=2%28-1%29-4
m=-2-4
m=-6

Substituting r=3 into

m=2r-4
m=2%283%29-4
m=6-4
m=2

We don't need to check m=-6 for that was
the value of m in part (a), and we knew 
that in that case the roots were not only 
reciprocals but they were also equal.

Checking m=2

x%5E2-4x%2B7=m%28x-1%29
x%5E2-4x%2B7=2%28x-1%29
x%5E2-4x%2B7=2x-2
x%5E2-6x%2B9=0
%28x-3%29%28x-3%29=0
x-3=0, x-3=0    
x=3  , x=3

So there are two answers to (c),

m=-6 and m=2.

Edwin