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

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TopNewestWe 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 \(x = -\frac{n}{m}\).

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

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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

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Good one!I haven't think of this

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

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Ha Ha,that's true!

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