Math: Algebra?

Favorite Answer

2x⁴ - 16x² - 2 = 0

2.(x⁴ - 8x² - 1) = 0

x⁴ - 8x² - 1 = 0

x⁴ - 8x² = 1

x⁴ - 8x² + 16 = 1 + 16

(x² - 4)² = 17

x² - 4 = ± √17

x² = 4 ± √17 → a square cannot be negative

x² = 4 + √17

x = ± √(4 + √17)...Show more

First, remove the common factors
==> x^4 - 8x^2 - 1= 0
If you look closely at this, you will see that it could be considered a quadratic equation using x^2 and the base, rather than just x

so, substitute u for x^2
==> u^2 - 8u - 1 = 0
now solve for u
u = (8 +/- sqrt(64 + 4)/2
==> u = (8 +/- sqrt(68)/2
==> u = 4 + sqrt(17) or 4 - sqrt(17)
x = sqrt(u)
==> x = +/- sqrt(4 + sqrt(17)) or +/- sqrt(4 - sqrt(17))


Note that sqrt(4 -sqrt(17)) is a complex number (sqrt(17)> 4)
so +/- sqrt(sqrt(17)-4)i where i = sqrt(-1)...Show more

2x^4 - 16x^2 - 2 = 0
2 (x^2 - 4)^2 - 34 = 0
Real solutions:
x = -sqrt(4 + sqrt(17))
x = sqrt(4 + sqrt(17))
Complex solutions:
x = -i sqrt(sqrt(17) - 4)
x = i sqrt(sqrt(17) - 4)...Show more

Presuming you want to solve for x:

2x⁴ - 16x² - 2 = 0

If we make this substitution:

z = x²

we get:

2z² - 16z - 2 = 0

We now have a quadratic that we can solve.  Let's simplify this by dividing both sides by 2, then I'll complete the square:

z² - 8z - 1 = 0
z² - 8z = 1
z² - 8z + 16 = 1 + 16
(z - 4)² = 17
z - 4 = ± √17
z = 4 ± √17

Now we can substitute back the expression in terms of x and solve for x:

x² = 4 ± √17
x = ± √(4 ± √17)

So we have four total roots.  Two of them are complex as 4 - √17 is negative.  The other two are real....Show more

what is the question?...Show more