Simultanious equations using quadratics

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Simonsky

Junior Member
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Jul 4, 2017
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Having a few issues with:

k(k-m) = 12
k(k+m) = 60

Not sure how to do this but will try, so:

expand: k^2 -km = 12 and k^2 + km = 60. So adjusting one equation to get a substitution for K^2 : k^2 = 12+km, so:

(12+km)(12+km) - km -12 = 0

km^2 + 23km + 132 = 0 and this (if I did it correctly) factorises to (mk+11)(mk+12) but now having mk =-11 and mk =-12 doesn't help.

Can someone nudge me in a better direction? Thanks
 
You claim that knowing what mk equals doesn't help. You also claim that k^2 = 12 + km. Since you know km why cant you solve for k??? Are you sure that you have the correct value for mk?


Here is what I would do
First see if k=0 works. If it does, then find the value for k and be done for this part.

2ndly, divide by k and get (k+m)/(k-m) = 5. Solve this for k in terms of m or visa versa and plug back in and solve for one variable.

See what you can do from here.
 
Thanks..not sure I follow what you say. I've checked my factorising which seems OK (but might not be). Still ot clear,,may need more nudging-sorry!
 
Actually, there is a simple way that happens to work in this case, and a general way that will work in any such case.

Obviously k cannot be zero so

[MATH]k(k - m) = 12 \implies (k - m) = \dfrac{12}{k} \implies \\ m = k - \dfrac{12}{k} = \dfrac{k^2 - 12}{k}.[/MATH][MATH]\therefore k(k + m) = 60 \implies k^2 + \dfrac{\cancel k(k^2 - 12)}{\cancel k} = 60 \implies \\ 2k^2 - 12 = 60 \implies \text {WHAT?}[/MATH]That is probably the way that I would do it: substitution is a general solution. But those with a better eye might "see"

[MATH]k(k - m) = 12 \implies k^2 - km = 12\\ k(k + m) = 60 \implies k^2 + km = 60\\ \therefore k^2 - km + k^2 + km = 12 + 60 \implies \text {WHAT?}[/MATH]Whichever way you get there, the rest is rather obvious.
 
Thanks Jeff, very elegantly done! I still haven't got that 'better eye.' Many thanks.
 
Simonsky said:
Having a few issues with:

k(k-m) = 12
k(k+m) = 60

Not sure how to do this but will try, so:

expand: k^2 -km = 12 and k^2 + km = 60. So adjusting one equation to get a substitution for K^2 : k^2 = 12+km, so:

(12+km)(12+km) - km -12 = 0

km^2 + 23km + 132 = 0 and this (if I did it correctly) factorises to (mk+11)(mk+12) but now having mk =-11 and mk =-12 doesn't help.

Can someone nudge me in a better direction? Thanks
Click to expand...
You took the 1st equation, distributed and then solved for k^2. You got k^2 = 12 + km. So far very good. But then you plugged the value for k^2 back into the same equation. Meaning you ignored the 2nd equation!

The 2nd equation is k(k+m) = k^2 + km =60. In this equation you should replace k^2 with 12+ km. So (12+km) + km = 60. Then 2km = 48 and km = 24.

(for the record if you had just computed (k^2+km) - (k^2-km) = 60-12 that too leads to km = 24)

We know that k^2 + km = 60, so k^2 + 24 = 60 and then k^2 = 36. Continue from here.
 
To repeat what I said above.

k(k+m) = 60 implies k^2 + mk = 60
k(k-m) = 12 implies that k^2 -km = 12

Just subtract the two equations and get 2km = 48. Then km = 24. Now follow other posts.
 
Simonsky said:
Having a few issues with:

k(k-m) = 12
k(k+m) = 60

Not sure how to do this but will try, so:

expand: k^2 -km = 12 and k^2 + km = 60. So adjusting one equation to get a substitution for K^2 : k^2 = 12+km, so:
Click to expand...
I would, rather, add the two equations to get 2k^2= 72. k^2= 36, k= 6. Then k^2- km= 36- 6m= 12 so -6m= -24, m= 4. As a check, put k= 6, m= 4 in k^2+ km= 60. k^2+ 6k= 36+ 24= 60 is correct

(12+km)(12+km) - km -12 = 0

km^2 + 23km + 132 = 0 and this (if I did it correctly) factorises to (mk+11)(mk+12) but now having mk =-11 and mk =-12 doesn't help.

Can someone nudge me in a better direction? Thanks
Click to expand...
 
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