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Find all integer values of $x$ where $x^2+3x+24$ is a perfect square. By guessing I found one solution that $x=5$.I found the problem in the exercise of a book in the chapter of polynomials. So please help me.

Sufaid Saleel
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    Hint: complete the square. Second hint: to avoid getting fractions, use the fact that if $x^2+3x+24$ is a square then so is $4(x^2+3x+24)$. – Erick Wong Dec 17 '16 at 05:22
  • More generally there's also this question (and probably some others; that's just the one I personally remember answering). – Micah Dec 17 '16 at 05:55

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$x^2+3x+24 = n^2$ so $x^2+3x+9/4 = n^2-24+9/4$ or $(x+3/2)^2 =n^2-87/4$ or $(2x+3)^2 =4n^2-87$.

Therefore $87 =4n^2-(2x+3)^2 =(2n-2x-3)(2n+2x+3) $.

$87=3*29$, so it can be factored as $87 =1*87,3*29 $.

Therefore $(2n-2x-3,2n+2x+3) =(-87, -1), (-29, -3), (3, 29), (1, 87) $.

If $(2n-2x-3,2n+2x+3) =(a, b) $, then $4x+6 = b-a$ and $4n=b+a$ so $x = (b-a-6)/4$ and $n = (b+a)/4$.

For these 4 sets, $x = (-87+1-6)/4 =-92/4 = -23 $, $x = (-29+3-6)/4 =-32/4 = -8 $, $x = (29-3-6)/4 =20/4 = 5 $, $x = (87-1-6)/4 =80/4 = 20 $.

marty cohen
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Since the giving term is larger than (x+1)^2 and is smaller than (x+5)^2, then it must equal to the square of (x+2) or (x+3) or (x+4), plug these 3 terms into the equation and you will get the answer.

Jason
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  • This is good advice for more general problems of this type such as high-degree "nearly square" polynomials, where exact factorization isn't possible but you can still reduce to a finite set of cases. – Erick Wong Dec 17 '16 at 13:29
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Dario Alpern has a nice calculator that will show you a step-by-step derivation of the four solutions, which are $x = -23,-8,5,20.$ You should input $a = 1, c = -1, d = 3, f = 24$ in the left frame.

user399937
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