\hspace{-16}$The no. of Integral values of $\mathbf{a}$ so that $\mathbf{x^2-(a+1)x+(a-1)=0}$\\\\ has Integral Roots.
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36eqn is x2 - (a + 1)x + (a - 1) = 0
D = a2 + 2a + 1 - 4 (a - 1) = a2 - 2a + 5 = (a - 1)2 + 4 > 0 for all a E R
thus, this eqn. will hv real and distinct roots for all a E R
Now, using ur method,
a (x - 1) = x2 - x - 1
or, a = x2 - x - 1x - 1 = x + 1 - xx - 1
Now, since, x is an integer then the only values of x for xx - 1 to be an integer are 0 and 2
setting x = 0 and 2 we get, a = 1 and 1
Thus, the only value of a is 1
nt sure used ur way... 'nakal mein bhi akal chahiye' :P
