-3x^4+27x^2+1200=0
-3(x^2+16)(x^2-25) = 0
That better?
thats the step i got to before i dont know what to do
what? Have you forgotten your Algebra I?
-3(x^2+16)(x-5)(x+5) = 0
the product if zero if any of the factors is zero.
im asking what the zeros for the equation are
the zeros of the function are where any factor is zero. What are you doing in pre-cal???
when is -3 = 0? never
when is x^2+16 = 0? never (x^2 is always positive, so x^2+16 cannot be zero)
x-5=0 when x=5
x+5=0 when x = -5
So, the zeros are x = ±5
Looks like you have some serious review to do.
-3x^4+27x^2+1200=0
0=3x^4-27x^2-1200
(3x²+48)(x²-25)=0
3(x²+16)(x+5)(x-5)=0
To find the roots of the equation -3x^4 + 27x^2 + 1200 = 0, we can use the quadratic formula. Although the equation is a quartic equation, we can treat it as a quadratic by substituting x^2 as a single variable, let's say y.
So, let y = x^2. Substituting this back into the equation, we get -3y^2 + 27y + 1200 = 0.
Now, we can solve this quadratic equation for y using the quadratic formula: y = (-b±√(b^2-4ac))/(2a), where a, b, and c are the coefficients of y.
For this equation, a = -3, b = 27, and c = 1200. Plugging these values into the quadratic formula, we get:
y = (-27±√(27^2 - 4(-3)(1200)))/(2(-3))
= (-27±√(729 + 14400))/(-6)
= (-27±√15129)/(-6)
= (-27±123)/(-6)
Now, we have two values for y: (-27+123)/(-6) = 96/(-6) = -16 and (-27-123)/(-6) = -150/(-6) = 25.
Remember, y = x^2. Substituting the values of y back into the equation, we get:
x^2 = -16 or x^2 = 25
To find the roots, take the square root of both sides:
For x^2 = -16, since square roots of negative numbers are not real, there are no real roots for this part of the equation.
For x^2 = 25, taking the square root of both sides gives x = ± 5.
Therefore, the roots of the original equation -3x^4 + 27x^2 + 1200 = 0 are x = 5, x = -5.