Question 202021: Could you help me about this?
For what value of K will the equation (2-k).x^2+(1-k).2x+(k-1) be a perfect square?
Found 3 solutions by Theo, ikleyn, Edwin McCravy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I think k = 1 should do it.
then all the variables cancel out except x^2.
equation becomes:
(2-1)*k^2 + (1-1)*2x + (1-1)
which becomes
k^2 + 0 + 0
which becomes
k^2
Answer by ikleyn(52914)  (Show Source):
You can put this solution on YOUR website! .
For what value of k will the  expression (2-k).x^2+(1-k).2x+(k-1) be a perfect square?
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Tutor @Theo guessed one solution: it is k = 1.
It is good when a person can correctly guess a solution.
But this problem has another solution, which is much more difficult to guess.
So, our duty and our task is to find all possible solutions.
Behind it, there is one beautiful idea, which will justify all our efforts.
The necessary and sufficient condition for this given quadratic polynomial to be
a perfect square is the condition that the discriminant of this quadratic polynomial
is 0 (zero).
The discriminant is d = b^2 - 4ac, referring to the general form of the quadratic polynomial.
In our case, a = (2-k), b = 2(1-k), c = (k-1).
So, we write for the discriminant
d = (2(1-k))^2 - 4*(2-k)*(k-1) = 0.
We see this common factor (k-1) in both terms, so we take it out of the expression
d = (1-k)*(4*(1-k) + 4*(2-k)) = (1-k)*(4 - 4k + 8 - 4k) = (1-k)*(12 - 8k).
First factor, (1-k) gives us one root k = 1.
It is the value guessed by @Theo.
Second factor gives us 12 - 8k = 0, or k =  =  .
Thus the second root of the discriminant equation is the second solution to the problem.
ANSWER. There are TWO and ONLY two values of 'k' making the given expression perfect square.
These values are k = 1 and k = 3/2.
Solved completely.
Nice problem, nice method and nice solution.
Good piece to learn ( ! )
My congratulations ( ! ) Enjoy ( ! )
Answer by Edwin McCravy(20065)  (Show Source):
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