what will be the value m if one zero of polynomial (m^2+4)x^2+63+4m is reciprocal of other
Answers
f(x) = (m^2+4)x^2+63+4m
Let y be the zero of the polynomial f(x)
The zero will be 1/y
On putting x=y in the polynomial f(x), we get
f(y) = y^2 × m^2 +4y^2 +63 +4m = 0 ...(i)
On putting x=1/y in the polynomial f(x), we get
f(1/y) = (1/y)^2 × m^2 +4(1/y)^2 +63 +4m =0 ...(ii)
On equating and canceling the like terms in the equation (i) and (ii) , we get
y^2 × m^2 +4y^2 = (1/y)^2 × m^2 + 4(1/y)^2
y^2 (m^2 + 4) = (1/y)^2 (m^2+4)
y^4 = 1
y = 1 or -1
On putting y = 1 in the equation (i), we get
(m^2 + 4)×(1)^2 +63 + 4m = 0
m^2 + 4 +63 + 4m =0
m^2 +4m + 67 = 0
The answers you get after solving this quadratic equation will be both complex.
So, the answers will be -2-3i√7 and -2+3i√7
Hope this will help you
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