Positive roots of equation x^2 + (4/x^2) = 5 (can't get ans)

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Find the positive roots of the equation:

x^2 + (4/x^2) = 5

I know it's 1 and 2 but can't prove it. I didn't know whether to try and solve it as x^2 + 4x^-2 - 5 = 0 or multiply both sides by x^2 i.e.:

x^4 + 4 = 5x^2
x^4 - 5x^2 = -4
x^2(x^2 - 5) = -4

x^2 = -4 or x^2 - 5 = -4 ???

But that's wrong...

Anyway, thanks for any help.
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Re: Positive roots of equation

Hello, Monkeyseat!

Evidently, you've forgotten some basic algebra . . .


Find the positive roots of the equation: .\(\displaystyle x^2 + \frac{4}{x^2} \:=\:5\)

\(\displaystyle \text{We have: }\;x^2 - 5 + \frac{4}{x^2}\:=\:0\)

\(\displaystyle \text{Multiply by }x^2\!:\;\;x^4 - 5x^2 + 4 \:=\:0\)

\(\displaystyle \text{Factor: }\;(x^2-1)(x^2-4) \:=\:0\)

\(\displaystyle \text{Factor: }\;\;(x-1)(x+1)(x-2)(x+2) \:=\:0\)

\(\displaystyle \text{Hence: }\;x \;=\;\pm1,\:\pm2\)


\(\displaystyle \text{The }positive\text{ roots are: }\:x \:=\:1,\:2\)

 

Loren

Senior Member
Joined
Aug 28, 2007
Messages
1,299
Re: Positive roots of equation

In solving an equation such as this, the procedure can be to get all terms on one side of the equation with zero on the other side. You then attempt to factor the non-zero expression so that you have factors that when multiplied = zero. This tells you that at least one of those factors is zero, because if none of them were zero the product could not be zero.
You applied this procedure to an equation with -4 instead of zero. In this case there are an infinite number of ways to multiply two factors together to get -4. You approached the problem as if there were only two ways. You have to get zero on one side of the equation and factors on the other side in order to employ that method.
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Much appreciated.

Thanks! :)
 
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