# X2 -little bit of help.Watch

#1
x^3-75x+250 = 0

Solve...

I'm stumped. Anyone?

thanks ^^
0
10 years ago
#2
Clue: -1000 + 750 + 250 = 0
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10 years ago
#3
Not sure if this is the best way but take the 250 to the other side and factor out x.

Can you see where to go from there?
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10 years ago
#4
(Original post by flimp)
Not sure if this is the best way but take the 250 to the other side and factor out x.

Can you see where to go from there?
I can't???
0
10 years ago
#5
To solve an equation like this, you have to find a number (which we'll call a) which, when x is replaced by a, makes the LHS of the equation equal to zero. It's mainly guesswork to be honest.

Once you've found a, then this shows the (x - a) is a factor of the LHS, so you can use long division to divide the LHS by (x - a), which will result in a quadratic to solve to obtain the other two roots.
10 years ago
#6
(Original post by flimp)
Not sure if this is the best way but take the 250 to the other side and factor out x.

Can you see where to go from there?
??
10 years ago
#7
(Original post by flimp)
Not sure if this is the best way but take the 250 to the other side and factor out x.

Can you see where to go from there?
That's not much use, unless you know x is an integer. And even then it's use is limited.
#8
I thought you did

x(x^2- 75) = 250

... and went from there. It's ages since I did stuff like this. >_<
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10 years ago
#9
(Original post by in_jeopardy)
I thought you did

x(x^2- 75) = 250

... and went from there. It's ages since I did stuff like this. >_<
That only makes it harder

Factor theorem. If F(a)=0 then (x-a) is a factor. Then use polynomial long division.
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#10
So, I have to guess at a number?
0
10 years ago
#11
(Original post by in_jeopardy)
So, I have to guess at a number?
Yep, and if you look up to notnek's post, that'll give you a clue.
10 years ago
#12
Trial and error pal.
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10 years ago
#13
yes, you start at (x-1) in this case x=1 then (x-1) then (x-3) mainly trial and error this is for the positive numbers of x
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10 years ago
#14
(Original post by Frater)
Trial and error pal.
But you may be able to spot it without using trial and error. I saw the 75 and the 250, thought 750+250=1000 and went from there.
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10 years ago
#15
(Original post by notnek)
But you may be able to spot it without using trial and error. I saw the 75 and the 250, thought 750+250=1000 and went from there.
Yeh, also, try to differentiate it. Then find the value of x.... that ll give you a value, then divide through by the that value to get a quadratic. then solve from there.... thats probably what they want to see, but I think notneks method is quicker, although you won't get the method marks.
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10 years ago
#16
(Original post by Frater)
Yeh, also, try to differentiate it. Then find the value of x.... that ll give you a value, then divide through by the that value to get a quadratic. then solve from there.... thats probably what they want to see, but I think notneks method is quicker, although you won't get the method marks.
Can you explain that method further? I may be being stupid but I don't see how differentiating will help.
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10 years ago
#17
ok, i apologize in advance, i am clueless with latx so I won't try

x^3 - 75x + 250 = 0

dy/dx = 3x^2 - 75

3x^2 = 75
x^2 = 25
x = +/- 5

Sub in to find -5 doesnt work so only +5 will

Divide x^3 -75x +250 by (x-5)

Answer is x^2 + 5x - 50

factorise (x - 5)(x+10)

so answer is (x - 5)^2 (x+10)

x= 5 or -10
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10 years ago
#18
Am I missing something?

You are solving dy/dx=0 to find the stationary points of the cubic function. Then you assume that one of the stationary points must be a root of the equation. Why?

Are you saying this would work for any equation?

e.g. x^2 + 2 = 0

dy/dx = 2x = 0
-> x = 0

But 0 is not a root of the equation...
0
10 years ago
#19
It's just a coincidence that one of the roots is also a turning point. Differentiation will not help at all, unless you want to sketch a graph to find the approximate location of the roots.
10 years ago
#20
notnek whats 28-3?

yes thats how many points ur beloved NUFC are from the drop
0
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