SOLUTION: Find values of n in the equation 2x^2 - 5(n-1)x +12=0 if the two roots are consecutive.

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Question 1118753: Find values of n in the equation
2x^2 - 5(n-1)x +12=0 if the
two roots are consecutive.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Looking at a few factorization trial combinations for the quadratic lead to
%282x%2B4%29%28x%2B3%29=2x%5E2%2B%28-1%29%2A5%28n-1%29x%2B12=00
and these roots are -3 and -2, which are consecutive integers.


Steps from that and then compare to the given quadratic expression:
%282x%2B4%29%28x%2B3%29
2x%5E2%2B4x%2B6x%2B12
2x%5E2%2B10x%2B12
from which corresponding parts give
-5%28n-1%29=10


Find n from that last equation.

Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find values of n in the equation
2x^2 - 5(n-1)x +12=0
if the two roots are consecutive highlight%28integer%29%29 highlight%28numbers%29.
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1.   Why you can not write the condition accurately and correctly ?   Why I should edit your writing ?


2.   The answer by @josgarithmetic in uncompleted.

        The complete answer is  THIS:

        There are two solutions.  One solution is:  the roots are  2  and  3,  and  n = 3.

        The other solution is:  the roots are  -3  and  -2,  and  n = -1.