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## 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

[tex]

E = \frac{Q}{4\pi\epsilon r^2}

[/tex]

## 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

[tex]

\frac{+4}{4\pi\epsilon . x^2} = \frac{+8}{4\pi\epsilon . (x-1)^2}

[/tex]

Simplifying:

[tex]

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

[/tex]

Square rooting & cross multiplying:

[tex]

\frac{\sqrt4}{x} = \frac{\sqrt8}{(x-1)}

[/tex]

[tex]

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

[/tex]

**It is at this point the textbook worked example says:**

Which simplifies to

[tex]

x = \frac{2}{(\sqrt8 + 2)}

[/tex]

**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:

[tex]

2x- 2 = \sqrt8(x)

[/tex]

And I can't get any further...

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