Basic manipulating equation algebra problem?

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  • #2
NascentOxygen
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How did this:
kx^(2) = 2mgh + 2mgx
Subtract 2mgh from both sides, then subtract 2mgx from both sides

i thought it's
kx^(2) = 2mgh + 2mgx
0 = 2mgh + 2mgx - kx^(2) ???
You can do it that way, too. Now multiply every term on both sides by -1
 
  • #3
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Subtract 2mgh from both sides, then subtract 2mgx from both sides


You can do it that way, too. Now multiply every term on both sides by -1
why do we need to mult both side by -1?
 
  • #4
NascentOxygen
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why do we need to mult both side by -1?
That gets you the equation that you were expecting to see, but by a different route.
 
  • #5
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That gets you the equation that you were expecting to see, but by a different route.


but what if you dont know the correct equation?

If i was doing this problem from scratch (without solution manuel)

I wouldnt have known this equation [0 = 2mgh + 2mgx - kx^(2)] was wrong because algebraically, i am correct.

is it like instincts from reading the problem to know what kind of equation to expect or something???
 
  • #6
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How did this:
kx^(2) = 2mgh + 2mgx
become this
kx^(2) - 2mgh - 2mgx = 0
Do you want to solve for x or what?
 
  • #7
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Do you want to solve for x or what?
yes.
 
  • #8
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yes.
Okay, now you got kx^(2) - 2mgh - 2mgx = 0 so you can use the quadratic formula to solve for x.
 
  • #9
NascentOxygen
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but what if you dont know the correct equation?

If i was doing this problem from scratch (without solution manuel)

I wouldnt have known this equation [0 = 2mgh + 2mgx - kx^(2)] was wrong because algebraically, i am correct.
It's not wrong. It's perfectly correct.

In general, it is customary to arrange the terms in decreasing degree, and with the leading term having a positive coefficient, e.g., x2 - 4x + 3 = 0

But this is just for appearances (and it makes it easier to comprehend), it's not wrong if you don't do this.

Multiplying both sides does not change anything material. If you do the same thing to both sides of an equation, you change nothing.
 

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