Finding a neutral point, electric fields. Simple algebra.

[SuperCup]

This is a question that I'm doing as part of revision for my January mocks. Being bad a algebra in general, I'm stuck. The answer is a worked example and I still don't understand it, so I'm feeling very silly.

Homework Statement

Charges of +4C and +8C are placed 1.00m apart. At what distance from the +4C charge is the electric field strength zero?

Homework Equations

$$E = \frac{Q}{4\pi\epsilon r^2}<br />$$

The Attempt at a Solution

I picture it like this:

<-----x-----><----------- 1-x ----------->
+4C -------- E=0 ----------------------- +8C

<----------------- 1m ----------------->

So, E due to 4C = E due to 8C
$$\frac{+4}{4\pi\epsilon . x^2} = \frac{+8}{4\pi\epsilon . (x-1)^2}$$

Simplifying:
$$\frac{4}{x^2} = \frac{8}{(x-1)^2}$$

Square rooting & cross multiplying:
$$\frac{\sqrt4}{x} = \frac{\sqrt8}{(x-1)}$$

$$2(x-1) = \sqrt8(x)$$


It is at this point the textbook worked example says:
Which simplifies to
$$x = \frac{2}{(\sqrt8 + 2)}$$
This is my problem, I don't know how on Earth it simplified to that. As the textbook hasn't explained the steps I feel silly as it's something probably really easy.

So I tried this:
$$2x- 2 = \sqrt8(x)$$
And I can't get any further...

I'd really appreciate a simple explanation, thankyou. :)

 

[RoyalCat]

Your mistake lies in the square root you took!
Notice that $$1-x$$ is positive by definition, while $$x-1$$ is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with $$1-x$$ in the denominator[/tex] instead of $$x-1$$ or just don't take the square root and solve the quadratic.

 

[SuperCup]

Your mistake lies in the square root you took!
Notice that $$1-x$$ is positive by definition, while $$x-1$$ is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with $$1-x$$ in the denominator[/tex] instead of $$x-1$$ or just don't take the square root and solve the quadratic.

Oops! My bad, I'll look out for that in future. Okay, so if I do it right, I get:
$$2(1-x) = \sqrt8(x)$$

And then
$$2 - 2x = \sqrt8(x)$$

... Uh, I still don't know how to go on from there to get $$x = \frac{2}{(\sqrt8 + 2)}$$, as the textbook says I should end up at.

The "solving the quadratic" bit seems to be my main problem. A giant chunk of GCSE Maths has escaped my memory haha. D: I feel rather daft as the Christmas holidays seem to have wiped my brain of how to do these!

 

[RoyalCat]

$$2 - 2x = \sqrt8(x)$$

Isolating all the $$x$$ terms on the left, and all the numbers on the right:

$$\sqrt{8}x+2x=2$$
$$(\sqrt{8}+2)x=2$$

Dividing both sides by $$\sqrt{8}+2$$ yields:

$$x=\frac{2}{\sqrt{8}+2}$$

Just some basic algebra, grab a book and get to practicing if you're rusty. ;)

 

[SuperCup]

$$2 - 2x = \sqrt8(x)$$

Isolating all the $$x$$ terms on the left, and all the numbers on the right:

$$\sqrt{8}x+2x=2$$
$$(\sqrt{8}+2)x=2$$

Dividing both sides by $$\sqrt{8}+2$$ yields:

$$x=\frac{2}{\sqrt{8}+2}$$

Just some basic algebra, grab a book and get to practicing if you're rusty. ;)

I can't believe this didn't occur to me. I'm feeling pretty silly :f Thankyou so much!