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if the root of the equation x²+px+q=0 differ from roots of equation x²+qx+p=0 by same quantity then A)p+q+1=0. B)q+p+2=0C)p+q+4=0. D)None Of These ?

if the root of the equation x²+px+q=0 differ from roots of equation x²+qx+p=0 by same quantity then A)p+q+1=0. B)q+p+2=0C)p+q+4=0. D)None Of These ?

Grade:11

2 Answers

Vikas TU
14149 Points
7 years ago
Let a and b be the roots for first quadratic and c and be the roots for second.
if the root of the equation x²+px+q=0 differ from roots of equation x²+qx+p=0
then,
a – b = c – d = k
Now difference of the roots in terms of sum of roots and prodduct of roots is:
=> |a-b| = root((a+b)^2 – 4ab))
and similarly,
|c-d| = root((c+d)^2 – 4cd))
putting the values we get,
root((-p)^2 – 4q)) – root((-q)^2 – 4p)) = k
Now k could be any assume it be zero
then,
root((-p)^2 – 4q)) – root((-q)^2 – 4p)) = 0
solving we get,
p + q + 4 = 0
 
mycroft holmes
272 Points
7 years ago
The given condition implies that the difference of the roots is the same for both quadratics.
 
Hence p^2-4q = q^2 - 4p \Rightarrow p^2 - q^2 = 4(p-q)
 
which is p, q are unequal leads to p+q+4 = 0

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