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July 24th, 2017, 04:24 PM   #1
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Unhappy 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
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July 24th, 2017, 04:37 PM   #2
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Not sure what you mean. Try p=4.
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July 24th, 2017, 05:29 PM   #3
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I dont know
I got this in a practice paper exactly like this.
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July 24th, 2017, 06:13 PM   #4
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$p x^2 + 15p + 4 = (20+3p)x$

$p x^2 -(20+3p)x + (15p+4) = 0$

$(a x + b)^2 = a^2 x^2 + 2 a b x + b^2$

$p = a^2$

$-(20+3p) = 2 a b$

$(15p+4) = b^2$

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.
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July 24th, 2017, 06:19 PM   #5
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Thanks to all!
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July 25th, 2017, 12:25 AM   #6
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What about P = 256?
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July 25th, 2017, 10:52 AM   #7
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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...
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July 25th, 2017, 12:30 PM   #8
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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.
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July 25th, 2017, 01:29 PM   #9
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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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