SOLUTION: If x2 - tx + t = 0, where t > 0, has one solution for x, then x is equal to A 1 B - 1 C - t + t2- t D 2 E - t

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Question 1187777: If x2 - tx + t = 0, where t > 0, has one solution for x, then x is equal to
A 1
B - 1
C - t + t2- t
D 2
E - t
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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If x2 - tx + t = 0, where t > 0, has one solution for x, then x is equal to
A) 1
B) - 1
C) - t + t2- t
D) 2
E) - t
~~~~~~~~~~~~~~~~~ 


We are given the equation  x^2 - tx + t = 0,  where t > 0,
and the problem asks to find its root x under the condition that this root is UNIQUE,
which means that there is no other roots.


It is the case, when a quadratic equation has MERGING roots.


For it, the necessary and sufficient condition is d = 0, where  " d "  is the discriminant 
of the quadratic equation  d = (b^2 - 4ac).



In our case,  b = -t, a = 1, c = t, so the discriminant is  d = (-t)^2 - 4t = t^2 - 4t.



The discriminant is zero,  t^2-4t = 0 = t*(t-4),  if  "t" is  0  or  4.



The case  t= 0  is excluded by the condition, since t must be positive.

So,  t = 4  is the only possibility.



For t = 4 we have

    x^2 -4t + 4 = (x-2)^2,


and the only root of it is  x = 2.


ANSWER.  Under given conditions,  x = 2.

Solved.