book says:

(4-4.00x + x^2)(6.00x10^-6C)= x^2(15.0x10^-6C)

they got:
3.00x^2+ 8.00x-8= 0

I don't get how they got that quadradic equation. (I mulitplied over and then divided the 15.0x10^-6 over and I didn't get what the book got)

(my problem that I'm doing is similar problem where

(d^2-dx+x^2)q1= x^2(3q)
would I be able to solve for x?
(I don't think so but since all they give is 3q, q, and d for a rod like the one below with a charged particle in equillibrium between the 2 charged outer particles. (one being 3q and other is q)

(3q)o==========o=====o(q)

total distance is given only as d.

Thank you

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3. 👁 61
1. first part reposted below

(4-4.00x + x^2)(6.00x10^-6C)= x^2(15.0x10^-6C)

3.00x^2+ 8.00x-8= 0 => how did they get this from the equation above?

I tried everything but I must be doing something incorrect.

Thank you.

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2. Divide both sides of the first equation by 6*10^-6C. Here's what I get:
(x^2 - 4x +4) = 2.5 x^2
Now double both sides to get rid of the fractional coefficient 2.5.
2x^2 - 8x +8 = 5x^2
3x^2 +8x -8 = 0
Use the quadratic equation to solve for x.
x = [-8 +/- sqrt 160]/6
= +0.7749 or -3.4415

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posted by drwls
3. divide both sides by 2x^(-6C) to get
(4 - 4x + x^2)(2) = 5x^2
8 - 8x + 2x^2 = 5x^2
rearrange:
3x^2 + 8x - 8 = 0

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posted by Reiny
4. Thank you both drwls and Reiny

I missed where they multiplied by 2 to get rid of the fraction.

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