Find all the zeroes of the equation. −2x4−64x2+288=0
To begin, let's substitute a variable to make the equation easier to work with. Let's substitute y = x².
The equation becomes: -2y² - 64y + 288 = 0.
Now, we can solve this quadratic equation for y. Factoring it, we have:
2y² + 64y - 288 = 0 ==> y² + 32y - 144 = 0.
We can factor this equation further:
(y + 36)(y - 4) = 0.
So, either y + 36 = 0 or y - 4 = 0.
If y + 36 = 0, then y = -36.
If y - 4 = 0, then y = 4.
Recall that y = x².
Therefore, either x² = -36 or x² = 4.
Solving these individually,
For x² = -36, taking the square root of both sides gives x = ±√(-36) = ±6i (imaginary numbers).
For x² = 4, taking the square root of both sides gives x = ±√4 = ±2.
So, the zeroes of the equation -2x^4 - 64x^2 + 288 = 0 are: x = ±6i and x = ±2.