? asked in Science & MathematicsMathematics · 2 months ago

# Math: Algebra?

2x^4-16x^2-2=0

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• 2 months ago

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)

• 2 months ago

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)

• ?
Lv 7
2 months ago

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)

• 2 months ago

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.