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Question

X^4 +15 = 8x^2

Answer 1

It is quite easy actually. It is a 'hidden' quadratic equation.
First, rearrange the terms: x^4-8x^2+15=0.
Then use substitution y=x^2
Thus we have: y^2-8y+15=0. Does it seems familiar to you?
solving it we will have: y=3 or 5
Thus x is +/-sqrt3 or +/-sqrt5.
Hope this helps:P

Answer 2

As with all quadratics set the equation equal to zero, thus:
x^4 - 8x^2 +15 = 0 resolving to
(x^2 - 3)(x^2 - 5) = 0 and solving each term
x^2 = 3 or x = sq rt of 3 = 1.732
and
x^2 = 0 or x = sq rt of 5 = 2.236

Answer 3

X^4 +15 = 8x^2
x^4 -8x^2 +15 = 0
(x^2 - 3)(x^2 - 5) = 0
x^2 - 3 = 0 | x^2 - 5 = 0
x = +/- sqrt 3 | x^2 = +/- sqrt 5

Answer 4

X^4 +15 = 8x^2

t=x^2

t^2+15 = 8t

t^2-8t+15=0

(t-5) * (t-3)=0

t=5 or 3

x=√5 or √3

Answer 5

you x^2 =y

so y^2+15=8y

or y^2-8y+15 =0

the roots are( 8+((64-60)^1/2))/2
or y=(8+2)/2 =5 and y=(8-2)/2 =3

do not forget to found x to take the square roots

so x= + square root 5 or x=- square root 5

and x= + square root 3 or x=- square root 3

so the roots are