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"Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.

In a quadratic in the form of $ax^2+bx+c$, the product of the roots of the quadratic equal the constant ($c$). Since there can only be integer coefficients, the only factors of $c$ ($11$) are $-1$ and $11$, and $1$ and $-11$. So we can put these in.

$(7x+1)(x-11)=$

$7x^2+x-77x-11=$

$7x^2-76x-11$

$N$ can be $-76$.

$(7x-1)(x+11)=$

$7x^2-x+77x-11=$

$7x^2+76x-11$

$N$ can be $76$. There is also another change you can make with the coefficient of $x^2$. You can move it over to the other side, so now we have....

$(x+1)(7x-11)=$

$7x^2+7x-11x-11=$

$7x^2-4x-11$

$N$ can be $-4$.

$(x-1)(7x+11)=$

$7x^2-7x+11x-11=$

$7x^2+4x-11.$

$N$ can be $4$.

So now we have these values of $n$: $-76$, $76$, $-4$, $4$.

Basically I am looking for a critique of my solution if it is correct.

Later
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user98535
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  • Yhe first sentence of the solution is wrong. In a quadratic $ax^2+bx+c$, the product of the roots is in fact $c/a$. – André Nicolas Oct 03 '13 at 19:05
  • i think your answer is correct – User8976 Oct 03 '13 at 18:59
  • Well, I think it is as well, but I was more looking for a critique of how I can my answer more... neat? Or just better I guess – user98535 Oct 03 '13 at 19:01
  • The question was IF the solution is correct, THEN please post a critique. Since you think that the solution is correct, what you wrote is not an answer, as it is not a critique of the solution at all. – Asaf Karagila Oct 04 '13 at 12:44

1 Answers1

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I think it is correct, but it would be easier to use that the discriminant must be a square number, i.e. $n^2+308$ must be a square number. This has the solutions $n=\pm 4, \pm 76$.

Dietrich Burde
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  • Sorry, this is a algebra class. I don't know what a discriminant is. – user98535 Oct 03 '13 at 18:58
  • Well, discriminant in usually taught in high school algebra... – DonAntonio Oct 03 '13 at 19:00
  • @user98535, for what it may be worth, I gave a detailed proof that square discriminant means the expression factors over the integers, not just over the rationals (with square discriminant), at http://math.stackexchange.com/questions/149820/how-to-factor-the-quadratic-polynomial-2x2-5xy-y2/149856#149856 For your case, keep the letter $x$ and just assign $y=1.$ – Will Jagy Oct 03 '13 at 19:09
  • @user98535 Proving it has exactly $4$ solutions without actually computing them is also an interesting excercise :) – chubakueno Oct 03 '13 at 19:28