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That question was Q->Show that the equation a/x-1 + b/x-2 + c/x-3 + d/ x-4 =1 can't have imaginary roots if a,b,c,d are any four real numbers of the same sign.

That question was Q->Show that the equation
a/x-1 + b/x-2 + c/x-3 + d/
x-4 =1 can't have imaginary
roots if a,b,c,d are any four
real numbers of the same
sign.

Grade:11

1 Answers

Shane Macguire
30 Points
10 years ago
i am reframing the solution may be this is helpful for the staudent

we will use the method of contradiction

let us ASSUME there is  a complex root p+iq of the given  equation  a/x-1 + b/x-2 + c/x-3 + d/x-4 =1

so put x = p + i q in the given equation then let us see the first term of the

a/(x-1) = a/(p+iq - 1) = a / {(p-1) + iq}

now multiply the numerator and denominator by (p-1) - iq

we get
 
a{(p-1) - i q}/{(p-1)^2 + q^2)}  -----------1

again we take the first term only

similarly we can  write  all the other  three terms

now we will put these terms in the given equation and write REAL PART and IMAGINARY PART seperately

imaginary part of L.H.S = imaginary part of R.H.S

and imaginary part of R.H.S = 0

so imaginary part of L.H.S is also 0

IGNORE the real part just concentrate on IMAGINARY PART of L.H.S, which will be

q( a/[(p-1)^2 + q^2)} + b/{(p-2)^2 + q^2)}+c/{(p-3)^2 + q^2)}+d/{(p-4)^2 + q^2)}]

there would be a term of q common so we took q as common

and we know this expression is equal to 0

so


q( a/[(p-1)^2 + q^2)} + b/{(p-2)^2 + q^2)}+c/{(p-3)^2 + q^2)}+d/{(p-4)^2 + q^2)}] = 0

let us take
( a/[(p-1)^2 + q^2)} + b/{(p-2)^2 + q^2)}+c/{(p-3)^2 + q^2)}+d/{(p-4)^2 + q^2)}= Y

now it of the form q * ( Y ) = 0

so either q = 0 or Y = 0
Y can''t be 0, it''s value depend on a,b,c,d, as the denominator is greater than 0 in all the terms ,as a,b,c,d are

either all +ve or -ve  value Y can''t be zero as if all are +ve then Y will be +ve if all -ve Y will be -ve

so only one case remaining q = 0

q is the IMAGINARY PART   so if imaginary part = 0   for all the cases where a,b,c,d are all +ve or -ve

so the root will be real it can''t be imaginary


imaginary part has to be 0 as a , b, c, d all are of same sign

the second  term will not be zero so q has to be zero
so our assumption was wrong there can''t be a imaginary root when a,b,c,d has
same sign

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