How to make mx+n=0 more complicated?

We all know how to solve the equation $mx+n=0$ ( $m,n$ are constants).The answer is of course $-\dfrac nm$ .

Then what if one day we forget about how to solve $ax+b=0$ ,but remember the quadratic formula for $ax^2+bx+c=0$ ? $x=\frac {-b \pm \sqrt {b^2-4ac}}{2a}$

I have 2 ways:

1. $mx+n=0,x(mx+n)=0$ .Solve for $mx^2+nx+0=0$ ,place $a\rightarrow m,b\rightarrow n,c\rightarrow0$ ,we get $x=\frac{-n\pm\sqrt{n^2-4m\times0}}{2m}=\frac{-n\pm|n|}{2m}=\frac{-n\pm n}{2m}=0\text{ or }-\frac nm$ That $0$ is of course not a root.So the root is $-\dfrac nm$

2.Solve for $0x^2+mx+n=0$ ,place $a\rightarrow0,b\rightarrow m,c\rightarrow n$ .

Wait, $a$ can't be $0$ ! Try using L'Hopital.

If $m>0$ ,then $\displaystyle\lim_{a\to 0}\frac{-m-\sqrt{m^2-4na}}{2a}$ doesn't exist. $\lim_{a\to 0}\frac{-m+\sqrt{m^2-4na}}{2a}=\lim_{a\to0}\frac{\dfrac{\partial}{\partial a}(-m+\sqrt{m^2-4na}) }{\dfrac{\partial}{\partial a}( 2a)}=\lim_{a\to0}\dfrac{\dfrac{-4n}{2\sqrt{m^2-4na}}}{2}=\frac{-n}{|m|}=-\frac nm$

If $m<0$ ,then $\displaystyle\lim_{a\to 0}\frac{-m+\sqrt{m^2-4na}}{2a}$ doesn't exist. $\lim_{a\to 0}\frac{-m-\sqrt{m^2-4na}}{2a}=\lim_{a\to0}\frac{\dfrac{\partial}{\partial a}(-m-\sqrt{m^2-4na}) }{\dfrac{\partial}{\partial a}( 2a)}=\lim_{a\to0}\dfrac{\dfrac{4n}{2\sqrt{m^2-4na}}}{2}=\frac{n}{|m|}=-\frac nm$

Can anyone think of other ways of solving $mx+n=0$ ?

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Comments

An alternate spin on your approach would be to start with $mx + n = 0$ and square both sides.

$m^2 x^2 + 2 m n x + n^2 = 0 \\ x = \frac{-2 m n \pm \sqrt{4 m^2 n^2 - 4 m^2 n^2}}{2 m^2} = -\frac{n}{m}$

The advantage of this is that we don't have to artificially neglect any roots

Steven Chase - 2 years, 11 months ago

Good one!I haven't think of this

X X - 2 years, 11 months ago

We could multiply both sides by $mx - n$ . Then $(mx + n)(mx - n) = 0(mx - n)$ becomes $m^2x^2 - n^2 = 0$ , and using the quadratic formula this solves to $x = \frac{-0 \pm \sqrt{0^2 - 4m^2(-n^2)}}{2m^2}$ or $x = \pm \frac{n}{m}$ . Then checking for extraneous solutions, we would eliminate $x = \frac{n}{m}$ and keep $x = -\frac{n}{m}$ .

David Vreken - 2 years, 11 months ago

Thank you.I haven't think of this one.

If we can remember how to graph, we could also just graph y = mx + n and find the x-intercept.

Ha Ha,that's true!

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