0
$\begingroup$

Hmm, I have been wondering about this when I went to solve the following equation: $$\frac{-x^2+2x}{5x-4} = 6$$

How come the above equation has two solutions, $-14 + 2\sqrt{55}$ and $-14 - 2\sqrt{55}$? I know when I simplify it, it turns to a quadratic equation, but how come it gets there? Also, does this apply to any rational equation in this form?

$\endgroup$
  • $\begingroup$ Do you know the formula for solving quadratic equations? $\endgroup$ – Capublanca Jan 27 '16 at 23:52
  • $\begingroup$ Yes, the value for $x$ in any equation in the form $ax^2 + bx + c = 0$ is $$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$$. $\endgroup$ – Obinna Nwakwue Jan 27 '16 at 23:56
  • 2
    $\begingroup$ Because it is an equation of the second degree. $\endgroup$ – Piquito Jan 28 '16 at 0:31
1
$\begingroup$

The equation can be rewritten as follows:

$$\frac{-x^2+2x-6(5x-4)}{5x-4} = 0 $$

Simplifying:

$$-x^2-28x+24=0$$

And solving:

$$x=-14\pm\sqrt{55}$$

$\endgroup$
  • $\begingroup$ I know that the solution is $x = -14 \pm \sqrt {55}$, but what is the triangular ellipses for? $\endgroup$ – Obinna Nwakwue Jan 28 '16 at 0:00
  • $\begingroup$ $\therefore$ means therefore $\endgroup$ – zz20s Jan 28 '16 at 0:04
  • $\begingroup$ It means therefore. You can ignore it if you like. In fact, I'll edit it out. $\endgroup$ – Lanier Freeman Jan 28 '16 at 0:42
3
$\begingroup$

In general, if $p(x)$ is a polynomial of degree $m$, $q(x)$ a polynomial of degree $n \ne m$ such that $p(x)$ and $q(x)$ are coprime as polynomials, and $c$ a nonzero constant, $\dfrac{p(x)}{q(x)} = c$ is equivalent to $p(x) - c q(x) = 0$, and since $p(x) - c q(x)$ has degree $\max(n,m)$, that is the number of solutions in $\mathbb C$ (counted by multiplicity).

$\endgroup$
  • $\begingroup$ So, let me clarify here. You are defining $p(x) = -x^2 + 2x$ and $q(x) = 5x - 4$, and $m$ is equal to the degree of $p(x)$, which is 2, while the degree of $q(x)$ is $1$ (which as defined as $n$), which satisfies the $n \neq m$ boolean, so $\frac {p(x)}{q(x)} = 6$ is equal to $(-x^2 + 2x) - 6(5x - 4) = 0$. Since that has degree of 2, that is the number of solutions! $\endgroup$ – Obinna Nwakwue Jan 28 '16 at 0:21
  • $\begingroup$ Wait a minute, why are there 2 solutions in the set of complex numbers? Don't you mean the set of real numbers? $\endgroup$ – Obinna Nwakwue Mar 5 '16 at 3:36
  • $\begingroup$ Okay, never mind, I get you. When dealing with quadratics, we work in the field of $\mathbb{C}$. $\endgroup$ – Obinna Nwakwue Jun 8 '16 at 17:01
0
$\begingroup$

$$ \frac{-x^2+2x}{5x-4} = 6 \iff \\ -x^2+2x = 6 (5x-4) \wedge 5x-4 \ne 0 \iff \\ x^2 + 28 x - 24 = 0 \wedge x \ne 4/5 $$ And a polynomial of order 2 has up to two real solutions.

$\endgroup$
  • $\begingroup$ I already know that; any polynomial with a degree of two has a maximum of two real solutions. By the way, what is the wedge sign for? $\endgroup$ – Obinna Nwakwue Jan 27 '16 at 23:58
  • $\begingroup$ It is a logical "and". $\endgroup$ – mvw Jan 27 '16 at 23:59
  • $\begingroup$ Note that, in MathJax, it can be also writen as \land (logical and). $\endgroup$ – JnxF Jan 28 '16 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.