Equations that end up being quadratic - substitution method

[mathwannabe]

Hello everybody

I got this equation. I can solve it brute forcing it, but it takes a long time... also, I am supposed to transform it into quadratic using substitution method, and solve it like that. Anyway, here is what I was trying to do before I got stuck:

1)

\(\displaystyle \dfrac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\)

I multiplied \(\displaystyle (x-1)(x-4), (x-2)(x-3)\) in the numerator, and \(\displaystyle (x+1)(x+4), (x+2)(x+3)\) in the denominator:

\(\displaystyle \dfrac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1\)

\(\displaystyle \dfrac{(x^2-5x+4)((x^2-5x+4)+2)}{(x^2+5x+4)((x^2+5x+4)+2)}=1\)

\(\displaystyle \dfrac{(x^2-5x+4)^2+2(x^2-5x+4)}{(x^2+5x+4)^2+2(x^2+5x+4)}=1\)

I have no idea how to proceed... or if I am on the right track, for that matter. For example, should I say \(\displaystyle t=x^2-5x+4\) and \(\displaystyle u=x^2+5x+4\)? Or should I first get rid of that 1 on the right side of the equation, in its current shape...

 

[Deleted member 4993]

Hello everybody

I got this equation. I can solve it brute forcing it, but it takes a long time... also, I am supposed to transform it into quadratic using substitution method, and solve it like that. Anyway, here is what I was trying to do before I got stuck:

1)

\(\displaystyle \dfrac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\)

I multiplied \(\displaystyle (x-1)(x-4), (x-2)(x-3)\) in the numerator, and \(\displaystyle (x+1)(x+4), (x+2)(x+3)\) in the denominator:

\(\displaystyle \dfrac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1\)

\(\displaystyle \dfrac{(x^2-5x+4)((x^2-5x+4)+2)}{(x^2+5x+4)((x^2+5x+4)+2)}=1\)

\(\displaystyle \dfrac{(x^2-5x+4)^2+2(x^2-5x+4)}{(x^2+5x+4)^2+2(x^2+5x+4)}=1\)

I have no idea how to proceed... or if I am on the right track, for that matter. For example, should I say \(\displaystyle t=x^2-5x+4\) and \(\displaystyle u=x^2+5x+4\)? Or should I first get rid of that 1 on the right side of the equation, in its current shape...

Let

x^2 - 5x = t and x^2 + 5x = s

then

(t+4)(t+6)/[(s+4)(s+6)] =1

t² + 10t + 24 = s² + 10s + 24

(t²-s²) +10(t-s) = 0

(t-s)(t+s+10) = 0

-10x(2x² + 10) = 0 .... and continue....

 

[mathwannabe]

Let

x^2 - 5x = t and x^2 + 5x = s

then

(t+4)(t+6)/[(s+4)(s+6)] =1

t² + 10t + 24 = s² + 10s + 24

(t²-s²) +10(t-s) = 0

(t-s)(t+s+10) = 0

-10x(2x² + 10) = 0 .... and continue....

Yes. I get it... Thank you

 

[Deleted member 4993]

Obvious that x = 0 is a solution: leaves 4*6 / (4*6) = 1

That is the only real solution.

 

[Deleted member 4993]

REALly ?

Would need power of IMAGINation for the other two - hockey-fans lack that :wink::wink: