Find all the zeroes of the equation.
- 2x^4 - 64x^2 + 288 = 0
To find the zeroes of the equation, we first need to substitute y=x^2 into the equation, so we get:
- 2y^2 - 64y + 288 = 0.
Now, we factor the quadratic equation:
- 2(y^2 - 32y + 144) = 0
- 2(y - 16)(y - 9) = 0.
Setting each factor to zero gives:
y - 16 = 0 -> y = 16
y - 9 = 0 -> y = 9.
Now we substitute back y=x^2 into the equation:
x^2 = 16 -> x = ±4,
x^2 = 9 -> x = ±3.
Therefore, the zeroes of the equation -2x^4 - 64x^2 + 288 = 0 are x = ±4 or ±3.