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July 24th, 2017, 03:24 PM  #1 
Newbie Joined: Jul 2017 From: New Zealand Posts: 7 Thanks: 0  Algebra Help  Perfect squares
The quadratic equation: px^2 +15p + 4 = (20 + 3p)x is a perfect square. Find the value(s) of p. Any help would be appreciated 
July 24th, 2017, 03:37 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,685 Thanks: 700 
Not sure what you mean. Try p=4.

July 24th, 2017, 04:29 PM  #3 
Newbie Joined: Jul 2017 From: New Zealand Posts: 7 Thanks: 0 
I dont know I got this in a practice paper exactly like this. 
July 24th, 2017, 05:13 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 
$p x^2 + 15p + 4 = (20+3p)x$ $p x^2 (20+3p)x + (15p+4) = 0$ $(a x + b)^2 = a^2 x^2 + 2 a b x + b^2$ $p = a^2$ $(20+3p) = 2 a b$ $(15p+4) = b^2$ probably the easiest way to solve this is just plug squared integers into the last equation $\left( \begin{array}{ccc} p &15p+4 &\sqrt{15p+4} \in \mathbb{N}\\ 1 & 19 & \text{False} \\ 4 & 64 & \text{True} \\ 9 & 139 & \text{False} \\ 16 & 244 & \text{False} \\ 25 & 379 & \text{False} \\ \end{array} \right)$ and as Denis said $p=4$ is the correct answer. 
July 24th, 2017, 05:19 PM  #5 
Newbie Joined: Jul 2017 From: New Zealand Posts: 7 Thanks: 0 
Thanks to all! 
July 24th, 2017, 11:25 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,050 Thanks: 1395 
What about P = 256?

July 25th, 2017, 09:52 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,685 Thanks: 700 
Not being sure, I solved for p to get discriminant sqrt(51p^2 + 104p + 400), which has only p=4 as integer solution. Apologies if I missed something... 
July 25th, 2017, 11:30 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,050 Thanks: 1395 
The question is poorly worded, as it's unclear what is meant by the equation being a perfect square. The equation can be rearranged so that one side is zero, but the question doesn't say this has to be done. If it is done, can one also multiply each side by a constant, so that p = 100/51 is equally acceptable? Arguably, that value is acceptable anyway.

July 25th, 2017, 12:29 PM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,685 Thanks: 700 
Turns out that x=4 and p=4 is only integer solution. Tested thru dumb loop code, each from 499 to 499. 

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algebra, perfect, squares 
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