# -3x^4+27x^2+1200=0

That better?

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## 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.

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## 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.

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## -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

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## 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.