Linear Equation to a Quadratic Equation Quick Question

doughishere

doughishere

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Dec 18, 2015
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Basically trying to solve for t using t*r=336 and (t+4)(r-2)=336.

(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= t*r=336
-2t2+4rt-8t = 0 <= multiply both sides by t, this makes the linear equation quadratic, right? Is this just do able because mathematics allows for multiplying both sides of the equation by the same thing.....could i go even further and make it -2t3+4rt2-8t2 = 0 just capriciously?
 
doughishere said:
Basically trying to solve for t using t*r=336 and (t+4)(r-2)=336.

(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= t*r=336
-2t2+4rt-8t = 0 <= multiply both sides by t, this makes the linear equation quadratic, right? - Not really. It will not assist in solving those two equations!!

Is this just do able because mathematics allows for multiplying both sides of the equation by the same thing.....could i go even further and make it -2t3+4rt2-8t2 = 0 just capriciously?

You can go on doing it as long as t <> 0 - but why?
Click to expand...

.
 
Subhotosh Khan said:
.
Click to expand...


ok so let me finish it out....from the start. This is the solution straight out of the gmat prep book.

Given: tr=336 and (t+4)(r-2)=336.


(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= tr=336
-2t2+4rt-8t = 0
-2t2+4(336)-8t = 0 <= tr=336
t2+4t-672=0
(t-24)(t+28)=0
t = 24 or -28


 
doughishere said:
Basically trying to solve for t using t*r=336 and (t+4)(r-2)=336.

(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= t*r=336
-2t2+4rt-8t = 0 <= multiply both sides by t, this makes the linear equation quadratic, right? Is this just do able because mathematics allows for multiplying both sides of the equation by the same thing.....could i go even further and make it -2t3+4rt2-8t2 = 0 just capriciously?
Click to expand...
As Subhotosh Khan, you can do so as long as you want and, since you are given t*r = 336, t is never zero. So what do you get for the two solutions to the problem?
 
doughishere said:
ok so let me finish it out....from the start. This is the solution straight out of the gmat prep book.

Given: tr=336 and (t+4)(r-2)=336.


(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= tr=336
-2t2+4rt-8t = 0
-2t2+4(336)-8t = 0 <= tr=336
t2+4t-672=0
(t-24)(t+28)=0
t = 24 or -28


Click to expand...

that produces r = 14 or -12
 
doughishere said:
Basically trying to solve for t using t*r=336 and (t+4)(r-2)=336.

(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= t*r=336
-2t2+4rt-8t = 0 <= multiply both sides by t, this makes the linear equation quadratic, right? Is this just do able because mathematics allows for multiplying both sides of the equation by the same thing.....could i go even further and make it -2t3+4rt2-8t2 = 0 just capriciously?
Click to expand...
Just to add something further: When you perform (possibly) irreversible operations [such as multiplying by zero], you always have the possibility of introducing spurious roots. In this case you have the equivalent of,
t2 + 4t - 672 = 0
If I were to multiple through by t again, I would have
t (t2 + 4t - 672) = t3 + 4 t2 - 672 t = 0
which introduces the spurious root of zero. That is, zero is a solution to the last equation but not to the original equation. You can recover the first equation but only if t is not zero since you can not divide by zero.

If you don't pay attention to these spurious roots, it can lead to interesting 'facts' such as 'all integers are equal': Consider the 'problem' of solving for t when t=1. Multiply through by t to get
t2 = t
so that t = 0 and 1 are solutions. So now divide by t to get
t = 1
and plug in the solution t=0 and we have
0 = 1
or 0+1=1=1+1=2 or
1 = 2
Continuing that way we have
0 = 1 = 2 = 3 = 4 = ...
so all integers are equal.
 
?

doughishere said:
ok so let me finish it out....from the start. This is the solution straight out of the gmat prep book.

Given: tr=336 and (t+4)(r-2)=336.


(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= tr=336
-2t2+4rt-8t = 0
-2t2+4(336)-8t = 0 <= tr=336
t2+4t-672=0
(t-24)(t+28)=0
t = 24 or -28
Click to expand...

That is an interesting way to eliminate r and get to the single variable equation, t2+4t-672=0, which can be solved by the quadratic equation. It seems idiosyncratic though, a very special case, what if the problem had been:


(t+4)(r-2) = 336
tr=337

Why would they use this technique, are they demonstrating some principle other then a shortcut? Still, very cool though, hmmm, I think this type of substitution can come into play when eliminating an indexing variable from some parametric equations.

 
Dale10101 said:
That is an interesting way to eliminate r and get to the single variable equation, t2+4t-672=0, which can be solved by the quadratic equation. It seems idiosyncratic though, a very special case, what if the problem had been:


(t+4)(r-2) = 336
tr=337

Why would they use this technique, are they demonstrating some principle other then a shortcut? Still, very cool though, hmmm, I think this type of substitution can come into play when eliminating an indexing variable from some parametric equations.
Click to expand...

(t+4)(r-2) = 336.........................tr=337

tr + 4r -2t - 8 = 336

4r - 2t - 7 = 0

4*337 - 2t2 - 7t = 0

2t2 + 7t - 1348 = 0

Now solve as before.......
 
doughishere said:
Basically trying to solve for t using t*r=336 and (t+4)(r-2)=336.

(t+4)(r-2) = 336
rt-2t+4r-8 = 336
-2t+4r-8 = 0 <= t*r=336
-2t2+4rt-8t = 0 <= multiply both sides by t, this makes the linear equation quadratic, right? Is this just do able because mathematics allows for multiplying both sides of the equation by the same thing.....could i go even further and make it -2t3+4rt2-8t2 = 0 just capriciously?
Click to expand...
Although there are much better ways to solve 'this quadratic' here is what I think the author wants.

tr=336 so r=336/t

Then (t+4)(r-2)=336 becomes (t+4)(336/t - 2)=336

Then 336 -2t + 1344/t -8 =336 or -2t + 1344/t -8=0

Multiply by t to get a quad in t: -2t^2 + 1344 -8t =0 .......
 
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