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I would like to know, step by step, how to solve the following equation. I already know the roots, but I would like an explanation. Thanks in advance. $$x^2+19x+48=m^2$$ I use $$m^2$$ to indicate perfect squares so the question is: how can i find the value of x that gives me a perfect squares as the result?

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    What are the roots? – 2017-02-04
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    What would you like explained? – 2017-02-04
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    So, you just want to know the quadratic formula? Or how to complete the square? – 2017-02-04
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    The question isn't really explicit. We have no idea what $m$ is, either a variable or constant. – 2017-02-04

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If $m$ is a constant, the roots are:

$$x=\frac{-19\pm\sqrt{19^2-4 (1)(48-m^2)}}{2(1)}=\frac{-19\pm\sqrt{169+m^2}}{2}$$

Since the discriminant is always positive, the equation always has at least one real root. How we proceed from here depends on what type of answer you want: real number, rational number, integer, natural number.

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    But m is not a constant. My question was intended to be: how can i find integer values of x, so x^2+19x+48 gives me a perfect square? – 2017-02-04
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    And I gave you a way to do so, assuming that $m$ is fixed for any particular equation considered. – 2017-02-04
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    Yes, but how do I find even M? If i gave you the polynomial " x^2+19x+48" and ask you find an x that gives a perfect square as the result, how would you procede to operate? – 2017-02-04
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    Treat $m$ as a parameter--a value that is held constant for any particular instance of the equation--and solve for $x$. – 2017-02-04
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I don't know what you mean by roots, since the roots are the solutions of $P(x)=0$ where $P$ is a polynomial.

So here, if you have the roots, you have the solutions.

To solve it, just subtract $m^2$ to both sides to get

$$x^2+19x+48-m^2=0$$

and this now a classic equation of degree $2$.

You can use the discriminant and well-know formulas to finish the resolution:

$$``x=\frac{-b\pm \sqrt \Delta}{2a}".$$

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    Is $m$ a variable or a constant in this problem? – 2017-02-04
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    @Oliver821 I think the OP want it to be a constant, so I treated it like it. – 2017-02-04
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    Yeah I was thinking the same thing. Just making sure is all. – 2017-02-04
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    I know I wasn't specific enough. Let me correct: how can i find a number x that gives me a perfect square? – 2017-02-04
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    @AlessandroFenu Than we understood you fine, and my answer works. – 2017-02-04
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    I am finding difficulties calculating the discriminant since I don't know m. Can you please show me a fast resolution to it? – 2017-02-04
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    @AlessandroFenu Your discriminant will necessarily depends on $m$. – 2017-02-04
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    Well this equation has the following solutions for x: 33, -3, -16, -52. How do i find them? Except for -3 and -16 which gives me 0 that is a perfect square, how do i find the other two possible values that gives me a perfect square? – 2017-02-04
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solve over $\mathbb R$

$x^2+19x+48-m^2=0$

We want $\Delta=19^2-4(48-m^2)=169+4m^2\ge 0$ to have real solutions.

This turns out to be always true, so solutions are given by $\{m\in \mathbb R, x=\frac{-19\pm\sqrt{169+m^2}}{2} \}$

solve over $\mathbb Z$ : diophantine equation

We want $\Delta$ to be a perfect square, that is $\delta^2=13^2+(2m)^2$

In addition to $m=0$, these are a pythagorician triplets, so let find them all :

$\begin{cases} u = r^2-s^2\\ v = 2rs\\ w = r^2+s^2 \end{cases}$

  • $v=13$ is not possible, since $13$ is odd.

  • $w=13=(\pm 3)^2+(\pm 2)^2$ is the only possibility, this give the triplet $(5,12,13)$ but this is not adequate for our case (because $\delta$ should be the greatest integer).

  • $u=(r-s)(r+s)=13$ since it is a prime, only $13\times 1$ is possible which gives $r=7, s=\pm 6$, which leads to the triplet $(13,84,85)$ and $m=\pm 42$.

Coming back to $x$, we have now :

  • $m=0$ , $\delta=\pm 13$ , $x=\frac{-19\pm 13}{2}=\{-16,-3\}$
  • $m=\pm 42$ , $\delta=\pm 85$ , $x=\frac{-19\pm 85}{2}=\{-52,33\}$

solve over $\mathbb Q$

$m=\frac pq$ with $p,q\neq 0$ (case $m=0$ already solved in $\mathbb Z$).

Equation $q^2x^2+19q^2x+48q^2-p^2=0$ has $\Delta=19^2q^4-4q^2(48q^2-p^2)=169q^4+4p^2q^2$

Let's call $a=13q$, $b=2p$, $\Delta=q^2\delta^2$ and again we must find pythagorician triplets $\delta^2=a^2+b^2$.

And for every triplet such that $13|a$ and $2|b$ you get rationnal solutions $x$.

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    Thanks. That was exactly what i was looking for. Love ya – 2017-02-04