How do I find two variables in algebra with a given equation?

AI Thread Summary
To find the values of p and x from the equations p = -x^2 - 2x + 100 and p = 8x + 25, the equations can be set equal to each other. This leads to the quadratic equation -x^2 - 10x = -75, which simplifies to x^2 + 10x = 75. The solution can be approached by factoring, completing the square, or using the quadratic formula. A correction was made regarding a miscalculation in the constant term, clarifying that it should be 25 instead of 75. The problem was ultimately resolved with the help of the forum participants.
GiTS
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Homework Statement


p=-x^2-2x+100 and p=8x+25, find p and x.


Homework Equations





The Attempt at a Solution


-x^2-2x+100=8x+125
-8x -8x
-100 -100
-x^2-10x=-75
x^2+10x=75
/10 /10
1/10x^2+x=75/10 i don't get how to get x alone on one side from here.
 
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Try (Ax^2) + Bx + C =0

then you know a formula to get x..
 
GiTS said:

Homework Statement


p=-x^2-2x+100 and p=8x+25, find p and x.


Homework Equations





The Attempt at a Solution


-x^2-2x+100=8x+125
-8x -8x
-100 -100
-x^2-10x=-75
x^2+10x=75
/10 /10
1/10x^2+x=75/10 i don't get how to get x alone on one side from here.

Oh, c'mon! That's a quadratic equation. Factor if you can, or complete the square or use the quadratic formula. Oh, by the way, 125- 100= 25, not 75.
 
HallsofIvy said:
Oh, c'mon! That's a quadratic equation. Factor if you can, or complete the square or use the quadratic formula. Oh, by the way, 125- 100= 25, not 75.

God I feel stupid. You'd think i'd recognize quadratic equations by now :P. I miswrote 125, supposed to be just 25. Thanks guys!
 
No Preblem!


Problem solved :P
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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