Question #7f145

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Question #7f145

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Answer 1 · 2017-12-29T23:11:48

$x = ⎧ ⎨ ⎩ \pm \sqrt{\frac{1 + i \sqrt{23}}{2}}, \pm \sqrt{\frac{1 - i \sqrt{23}}{2}} ⎫ ⎬ ⎭$ $color(red)( x= { pm sqrt(( 1 + isqrt(23) )/2) , pm sqrt(( 1 - isqrt(23) )/2)}$

Explanation:

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

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

$ψ = x^{2}$ $psi = x^2$

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

$\Rightarrow ψ^{2} = x^{4}$ $=> psi^2 = x^4$

Hence our original function transforms into:

$ψ^{2} - ψ + 6 = 0$ $psi^2 - psi + 6 = 0$

Now we can just solve like we ussually do:

We can just use the quadratic equation:

$ψ = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $psi = (-b pm sqrt( b^2-4ac) ) /(2a)$

For $a ψ^{2} + b ψ + c = 0$ $apsi^2 + bpsi + c = 0$

$ψ = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - (4 \cdot 1 \cdot 6)}}{2 \cdot 1}$ $psi = (-(-1) pm sqrt( (-1)^2 - (4*1*6) ) )/(2*1)$

$\Rightarrow ψ = \frac{1 \pm \sqrt{- 23}}{2}$ $=> psi = (1 pm sqrt(-23) )/2$

We know form our complex number studies that $i = \sqrt{- 1}$ $i = sqrt(-1)$

$\Rightarrow ψ = x^{2} = \frac{1 \pm i \sqrt{23}}{2}$ $=> psi = x^2 = ( 1 pm isqrt(23) ) /2$

$\Rightarrow x = \pm \sqrt{\frac{1 \pm i \sqrt{23}}{2}}$ $=> x =pm sqrt (( 1 pm isqrt(23) ) /2 )$

$\Rightarrow x = ⎧ ⎨ ⎩ \pm \sqrt{\frac{1 + i \sqrt{23}}{2}}, \pm \sqrt{\frac{1 - i \sqrt{23}}{2}} ⎫ ⎬ ⎭$ $color(red)(=> x= { pm sqrt(( 1 + isqrt(23) )/2) , pm sqrt(( 1 - isqrt(23) )/2)}$

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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 · 2017-12-29T23:15:56

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

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Question #7f145

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