# Solving the algebraic equation

I am trying to solve this: $$x-40={-400\over x}$$ The answer must be $x=20$
Please give step by step explanation.

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No, I did not give the full equation here .The actual problem is $$x={{(2{(x-10))}}{({({(x-5)}-{(x-10)})}{({({(x-5)}-{({(x-10)}-1})})})}\over x}$$ – Govind Balaji Mar 28 '13 at 5:34

Assuming $x\neq 0$, multiply both sides of the equation by $x$ to get: $$x\cdot(x-40) =-400$$ Next expand the left hand side and re-arrange to get: $$x^2-40x+400=0$$. To factor the left hand side, look for two numbers that when you multiply you get 400 and add to -40. Doing that yields $$(x-20)(x-20) =0.$$ You should be able to take it from here.

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I am not clear about how $x^2-40x+400$ becomes $(x-20)(x-20)$.Please explain it too. – Govind Balaji Mar 28 '13 at 5:42
I thought I made it clear in my answer. anyways, you could use the quadratic formula or recall that $(a+b)^2=a^2+2ab+b^2$ or in particular $(a-b)^2=(a-b)(a-b)=a^2-2ab+b^2$ with $a=x$ and $b=20$ – Nana Mar 28 '13 at 5:45

Step 1. Multiply through by x.

Step 2. Move everything to one side.

Step 3. Use the quadratic formula.

Step 4. Conclude that $x=20$.

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Please show your way by including the calculation process, for each step – Govind Balaji Mar 28 '13 at 5:27

Simplifying the equation gives :

$x^2 - 40x + 400 = 0$

$=> {(x-20)}^2 = 0$

$=> x = 20$

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