The product of two consecutive positive integers is 30 6 we need to find the integers
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let the two +ve consecutive integers be : n and n+1.
ATQ: n*(n+1)= 306
solving this:-
n^2 + n = 306
n^2 + n - 306 = 0
clearly, a=1, b=1, c=-306
therefore, putting these values in the following equation you will surely get the value of n and hence the numbers -
[ n = (-b±√b^2 - 4ac)/ 2a. ]
ATQ: n*(n+1)= 306
solving this:-
n^2 + n = 306
n^2 + n - 306 = 0
clearly, a=1, b=1, c=-306
therefore, putting these values in the following equation you will surely get the value of n and hence the numbers -
[ n = (-b±√b^2 - 4ac)/ 2a. ]
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