How to make mx+n=0 more complicated?

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

Then what if one day we forget about how to solve ax+b=0ax+b=0,but remember the quadratic formula for ax2+bx+c=0ax^2+bx+c=0?x=b±b24ac2ax=\frac {-b \pm \sqrt {b^2-4ac}}{2a}

I have 2 ways:

1.mx+n=0,x(mx+n)=0mx+n=0,x(mx+n)=0.Solve for mx2+nx+0=0mx^2+nx+0=0,place am,bn,c0a\rightarrow m,b\rightarrow n,c\rightarrow0,we get x=n±n24m×02m=n±n2m=n±n2m=0 or nmx=\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 00 is of course not a root.So the root is nm-\dfrac nm

2.Solve for 0x2+mx+n=00x^2+mx+n=0,place a0,bm,cna\rightarrow0,b\rightarrow m,c\rightarrow n.

Wait,aa can't be 00 ! Try using L'Hopital.

If m>0m>0,then lima0mm24na2a\displaystyle\lim_{a\to 0}\frac{-m-\sqrt{m^2-4na}}{2a} doesn't exist. lima0m+m24na2a=lima0a(m+m24na)a(2a)=lima04n2m24na2=nm=nm\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<0m<0,then lima0m+m24na2a\displaystyle\lim_{a\to 0}\frac{-m+\sqrt{m^2-4na}}{2a} doesn't exist. lima0mm24na2a=lima0a(mm24na)a(2a)=lima04n2m24na2=nm=nm\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=0mx+n=0?

Note by X X
1 year, 3 months ago

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

David Vreken - 1 year, 3 months ago

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Ha Ha,that's true!

X X - 1 year, 3 months ago

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We could multiply both sides by mxnmx - n. Then (mx+n)(mxn)=0(mxn)(mx + n)(mx - n) = 0(mx - n) becomes m2x2n2=0m^2x^2 - n^2 = 0, and using the quadratic formula this solves to x=0±024m2(n2)2m2x = \frac{-0 \pm \sqrt{0^2 - 4m^2(-n^2)}}{2m^2} or x=±nmx = \pm \frac{n}{m}. Then checking for extraneous solutions, we would eliminate x=nmx = \frac{n}{m} and keep x=nmx = -\frac{n}{m}.

David Vreken - 1 year, 3 months ago

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

X X - 1 year, 3 months ago

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

m2x2+2mnx+n2=0x=2mn±4m2n24m2n22m2=nm 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 - 1 year, 3 months ago

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Good one!I haven't think of this

X X - 1 year, 3 months ago

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