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the difference bt the roots of equaion (t+3)2+9(t-1)+k=0 is 5..then value of k is
(t+3)2 + 9(t-1) + k=0
=> t2 + 6t + 9 + 9t - 9 + k=0
=>t2 + 15t + k=0
let the roots of the above quad.eq be h and k.
sum of roots h+k = -b/a
= -15/1
= -15 . .......(eq 1)
difference of roots h - k = 5 (given).......(eq 2)
solving eq1 and eq2,
we get h=-5 and k=-10
product of roots h*k = c/a
=> (-5 X -10) = k/1
=> k=50
I will present another method of solving the problem
(difference between roots)2=(sum of roots)2-4(product of roots)
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