Question

Question #7f145

Answer 1

`color(red)( x= { pm sqrt(( 1 + isqrt(23) )/2) , pm sqrt(( 1 - isqrt(23) )/2)}`

Explanation:

This question can be solved via treating the function `x^4 - x^2 +6` as a quadratic...

The first thing we can do is make a simple substitution:

`psi = x^2`

Where `psi` is just a random variable i have choosen, you choose what ever you like

`=> psi^2 = x^4`

Hence our original function transforms into:

`psi^2 - psi + 6 = 0`

Now we can just solve like we ussually do:

We can just use the quadratic equation:

`psi = (-b pm sqrt( b^2-4ac) ) /(2a)`

For `apsi^2 + bpsi + c = 0`

`psi = (-(-1) pm sqrt( (-1)^2 - (4*1*6) ) )/(2*1)`

`=> psi = (1 pm sqrt(-23) )/2`

We know form our complex number studies that `i = sqrt(-1)`

`=> psi = x^2 = ( 1 pm isqrt(23) ) /2`

`=> x =pm sqrt (( 1 pm isqrt(23) ) /2 )`

`color(red)(=> x= { pm sqrt(( 1 + isqrt(23) )/2) , pm sqrt(( 1 - isqrt(23) )/2)}`

We see by a sketch that there are no solutions for `x in RR` , or in other words, there are only solutions in the complex relm

Answer 2

Expansion...

Explanation:

If `y = ax^(2n) + bx^n + c`

Then we can make a substitution:

`phi = x^n`

`=> phi^2 = (x^n)^2 = x^(2n)`

Hence our equation can be transformed into:

`y = aphi^2 + bphi + c`

What we know is much more simpler to deal with