In the equation:- y = 4 x 2 + a x + 2 5 Find out the minimum possible possitive natural number value of a such that for any natural number value of x ,value of y obtained is always a perfect square
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
This solution assumes that "If a polynomial achieves a perfect square for every integer value, then it must be a perfect square polynomial". Why must this be true?
Obtaining a perfect square trinomial on the RHS of the equation will make y = (ax + b)^2, which by definition will cause y to be a square number for any integer x, and therefore for any natural number x. Perhaps I'm looking at things wrong because of the lesser restriction that x need only be a natural number, but I'm not aware of any other way to guarantee that a second-order polynomial f(x) produce square numbers for every natural (integer) x without f(x) being a perfect square trinomial.
Reading the comment you posed to both solutions has led me to think more about the solution.
What you are looking for is a second-order polynomial f(x) = ax^2 + bx + c such that sqrt( f(x) ) = px + q. x is an integer (natural number) in both cases.
But if sqrt( f(x) ) = px + q, then by definition f(x) = (px + q)^2, which is always a perfect square trinomial.
As I observed below, I am trying to imagine a case where ax^2 + bx + c is always a square number without in turn being equal to (px + q)^2. I consider myself to have a good imagination, but I don't think it's quite that good.
In order for y to always be a perfect square,then the polynomial must be a perfect square trinomial. Observe that it can be written as : ( 2 x ) 2 + a x + 5 2 So the middle term must be 2 ( 2 x ) ( 5 ) = 2 0 x → a = 2 0
if y has to be a perfect square then when we factorize the p(x), it should also be a perfect square. so, using middle term factorizstion, we get 100x^2 = (10x)(10x) = ax/2 *ax/2. so, a/2 = 10. so a = 20.
Why must the factorization be a perfect square? Why can't it just attain perfect square values coincidentally?
Log in to reply
It may obtain perfect square values coincidentally but it is said that for which value of a will y always be a perfect square.In order for that to happen the polynomial must be a perfect square.
Why you wrote 100x^2=ax^2*ax^2
Log in to reply
aman,i'm sorry i little upset to you. i have found the value of a is 20 and -20 but the problem you write told me to find the value of x. who now this your false in write problems or this my bad in english. why not use indonesian language? hehe,im just kiding
Problem Loading...
Note Loading...
Set Loading...
In a perfect square trinomial, the discriminant of the quadratic formula b^2 - 4ac must equal 0.
This gives a^2 - 4(4)(25) = 0, which simplifies to a^2 - 400 = 0. This solves as a = {+20, -20}, and we reject -20 since it is not positive.