# why does the equation $(-x^2 + 2x)/(5x - 4) = 6$ have 2 solutions?

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?

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

## 3 Answers

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:

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

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

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).

• 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! – Obinna Nwakwue Jan 28 '16 at 0:21
• Wait a minute, why are there 2 solutions in the set of complex numbers? Don't you mean the set of real numbers? – Obinna Nwakwue Mar 5 '16 at 3:36
• Okay, never mind, I get you. When dealing with quadratics, we work in the field of $\mathbb{C}$. – Obinna Nwakwue Jun 8 '16 at 17:01

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

• 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? – Obinna Nwakwue Jan 27 '16 at 23:58
• It is a logical "and". – mvw Jan 27 '16 at 23:59
• Note that, in MathJax, it can be also writen as \land (logical and). – JnxF Jan 28 '16 at 0:55