If one zero of the quadratic polynomial (k^2+k)x^2+68x+6k is reciprocal of the other find k
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Solution:
Let m , 1/m are two zeroes of given
quadratic expression ,
Compare (k²+k)x²+68x+6k , with
ax²+bx+c , we get
a = k² + k = k(k+1)
b = 68 ,
c = 6k
Now ,
product of the zeroes = c/a
=> m × 1/m = ( 6k )/[k(k+1)]
=> 1= 6/(k+1)
=> k+1 = 6
=> k = 6 - 1
= k = 5
••••
Let m , 1/m are two zeroes of given
quadratic expression ,
Compare (k²+k)x²+68x+6k , with
ax²+bx+c , we get
a = k² + k = k(k+1)
b = 68 ,
c = 6k
Now ,
product of the zeroes = c/a
=> m × 1/m = ( 6k )/[k(k+1)]
=> 1= 6/(k+1)
=> k+1 = 6
=> k = 6 - 1
= k = 5
••••
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