Are you smarter than me? 36
Find the number of real of solutions of
x − 6 x 6 − 3 6 x 5 + 5 4 0 x 4 − 4 3 2 0 x 3 + 1 9 4 4 0 x 2 − 4 6 6 5 6 x + 4 6 6 5 6 = 0 .
The answer is 0.
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3 solutions
What do you mean by clearly. It's very difficult to analyse this.
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D o m a i n o f g ( x ) f ( x ) = D o m a i n f ( x ) ∩ D o m a i n g ( x ) , g ( x ) = 0
we are treating x − 6 ( x − 6 ) 6 as a different function , the graph of it would also be a curve but by definition of function it would be undefined at x =6 .
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You didn't got it right. I was saying about binomial expansion of (x-6)^6 . It is not easy to observe that above expression is binomial expansion of (x-6)^6.
I tried to use Descartes rule of sign, by virtue of which I got it can have at most 6 positive roots and no negative roots, but this gave me no beneficial information.
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@Prakash Chandra Rai – Sorry for providing wrong information, yes that's difficult to analyze , I saw x-6 in the denominator and applied division algorithm and got (x-6)^6
@Prakash Chandra Rai – Umm..I used Descarte's rule of sign change as well. Please tell if my solution is legitimate.
We know that, according to Descarte's rule of sign change, the number of times the sign changes in a polynomial, gives the total number of real roots of that polynomial. So, in the given expression:
x 6 − 3 6 x 5 + 5 4 0 x 4 − 4 3 2 0 x 3 + 1 9 4 4 0 x 2 − 4 6 6 5 6 x + 4 6 6 5 6
The original sequence of positive and negative signs is, + − + − + − +
By putting the value of x = − 1 , The signs are changed to + + + + + +
Clearly, in the second sequence there is no alternate sign view, and as a result, it can be concluded that this particular expression has no real roots.
A simple reason why you're wrong is the fact that x = 6 is a root of the expression. In Descartes' Rule of Signs , We check the number of sign changes in BOTH f ( x ) and f ( − x ) . Clearly, there are 6 sign changes in f ( x ) and 0 sign changes in f ( − x ) . The maximum number of positive roots is given by the number of sign changes in f ( x ) and the maximum number of negative roots is given by the number of sign changes in f ( − x ) .
This implies there are max 6 positive roots and 0 Negative Roots, not , as you claimed, no real roots.
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@Siddhartha Srivastava Thanks for the clarification! Then do we solve this question just by contracting the numerator in its binomial form???
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I think that is the only way. I didn't solve the problem and came here to see if there were any other alternate solutions.
Exactly we get that there are 4 possibilities of positive roots i.e number of positive roots can be 6 , 4 , 2 , 0 and the possibilities of negative root is 0 , but how have we proceeded to 0 real roots ??
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Everything you are saying is correct except the last line. We can't conclude that there are 0 real roots here mostly because the polynomial does have real roots.
My original comment was pointing out that the above solution is/was wrong. You can look Mehul's solution if you want to see the correct solution.
The expression is (x-6)^6 which gives one real solution x=6 but dividing this expression by x-6 results in 0/0 which is trivial and does not have any real solution. Therefore your logic is not right.
The question is poorly formulated, I guessed the answer because there were any solutions and the answer must be an integer but it its obviously that x=0 is not an imaginary solution that was my first thought.
We can clearly see that ( x − 6 ) 6 = x 6 − 3 6 x 5 + 5 4 0 x 4 − 4 3 2 0 x 3 + 1 9 4 4 0 x 2 − 4 6 6 5 6 x + 4 6 6 5 6 So the expression can be rewritten as
x − 6 ( x − 6 ) 6 = 0 so x could be 6
But wait It can't be 6 as it would result denominator as 0
∴ it has 0 Solutions