Finding the Perfect Value of k for a Square Trinomial

  • Thread starter Thread starter soggybread
  • Start date Start date
  • Tags Tags
    Square
AI Thread Summary
To determine the value of k that makes the expression 25x^2 - 60xy + ky^2 a perfect square trinomial, it can be expressed in the form (A + B)², where A² = 25x² and 2AB = -60xy. From A², A is identified as 5x, leading to the equation 2(5x)(B) = -60xy, which simplifies to B = -6y. Substituting B into the equation B² = ky² gives k = 36. This method confirms that the trinomial can indeed be expressed as a perfect square.
soggybread
Messages
20
Reaction score
0
ok, I'm pretty sure this is ridiculously simple, but I am stumped by this question..

What value of k would make 25x^2 - 60xy + ky^2 a perfect square trinomial?

I just don't know where to begin in order to find k. :rolleyes:

Any help would be much appreciated!
 
Physics news on Phys.org
soggybread said:
ok, I'm pretty sure this is ridiculously simple, but I am stumped by this question..

What value of k would make 25x^2 - 60xy + ky^2 a perfect square trinomial?

I just don't know where to begin in order to find k. :rolleyes:

Any help would be much appreciated!

Ok well starting with a binomial (A + B)
then (A + B)2 = (A + B)(A + B) = A2 + AB + AB + B2

which is the same as

A2 + 2AB + B2

Which would be a perfect square trinomial so do you think you could right the trinomial in question in this form for some value of k?
 
Just a tiny bit more help: as d leet said, a perfect square must be of the form (A+ B)2= A2+ 2AB+ B2.
Your formula is 25x2- 60xy+ ky2. That means you must have A2= 25x2, 2AB= -60xy and B2= ky2. From the first equation it should be obvious what A. Use that in the second equation to determine B and then use the third equation to find k.
 
Easy method - This is a perfect square. So discriminant must be 0. Put y = 1, so that this becomes a quadratic in x if you don't understand to usewhat I am talking about in the first statement. And I don't think this is a bad method.
So you would get 60^2 = 4*25*k.
Moreover here the value of y does not matter if you look deep. The co-efficient of first power of x is 60y and the constant is ky^2
So (60y)^2 = 4*25*k*y^2.
 
Ah! Thank You Guys! I see how it works now! :rolleyes:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top