I need to find the zeros in this equation.
-3x^4 + 27x^2 _ 1200 = 0

Question

I need to find the zeros in this equation. 
-3x^4 + 27x^2 _ 1200 = 0

michele_laino · Answer

please change the varible to a new variable, say y=x^2. Your eqution will become as below:
$$-3y ^{2}+27 y-1200=0$$
which can be solved easily for y

anonymous · Answer

I'm sorry, I'm not sure I understand. How were you able to get rid of a portion of the exponents just by changing the variable?

anonymous · Answer

Is this correct? $$-3x^4 +27^2 - 1200 + 1200 = 0 + 1200$$

anonymous · Answer

$$\sqrt{-3x^4 + 27^2} = \sqrt{1200}$$

michele_laino · Answer

please you have to apply the standard formula to solve the quadratic equation for y

anonymous · Answer

That first one was supposed to go second.

anonymous · Answer

What would be the standard formula again?

anonymous · Answer

I'm not trying to get an answer out of you I just need some help wrapping my head around this stuff.

michele_laino · Answer

$$ay ^{2}+by+c=0$$
then the standard formula is:
$$y=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

anonymous · Answer

Wait, you mean the quadratic formula? Okay, I get it now. But doesn't that help me find only the imaginary zeroes?

campbell_st · Answer

just looking at the dicriminant its negative so the zeros are complex

michele_laino · Answer

@ericdegr8  your equation is:
$$3y ^{2}-27y+1200=0$$
so?

anonymous · Answer

$$x = \frac{ 27\pm \sqrt{27^{2}-4(3)(-1200)} }{ 2(3) }$$

anonymous · Answer

$$x = \frac{ 27\pm \sqrt{729 + 14,400} }{ 6 }$$

anonymous · Answer

Correct?

campbell_st · Answer

I think if you are using the general quadratic formula on the original equation, you have a couple of errors

$$x^2 = \frac{-27 \pm \sqrt{(-27)^2 - 4 \times (-3) \times (-1200)}}{2 \times (-3)}$$

campbell_st · Answer

opps should be 27 inside the square root.. 

no the interesting problem is that you will now have to take the square root of the answer to find x

anonymous · Answer

Okay, so I use the quadratic formula to find x^2, then I find the square root of that?

anonymous · Answer

But what about the real solutions? I thought the quadratic formula only finds complex solutions? Don't you have to factor normally to find the real solutions?

anonymous · Answer

Or are there any real solutions?

campbell_st · Answer

the quadratic formula will find all solutions... 

but checking the discriminant when I 1st saw the question, it was negative so the solutions for x^2 will be complex... 

the discriminant is  $$\Delta = b^2 - 4ac$$

anonymous · Answer

Okay, so that's the discriminant. This actually clears up a lot of confusion in my math class.

anonymous · Answer

How about you @magepker728 ? Do you have any input?

campbell_st · Answer

can I just check the equation is 

$$-3x^4 + 27x^2 - 1200 = 0$$

asyou seem to have a typo before the constant term..

anonymous · Answer

Quick question: once I find the square root for the expression inside the sqrt symbol (b - 4ac), do I divide first, or split the equation into two, each with their respective + and - signs?

campbell_st · Answer

no... since it will be negative.... use i^2 = -1   to help make it a positive value 

e.g. $$\sqrt{-15} = \sqrt{15i^2} = i \sqrt{15}$$

campbell_st · Answer

so is the equation the 1 I typed above..?

anonymous · Answer

Oh, sorry, I think I made a mistake. In the equation, 1200 is added, so it looks like this:$$-3x^4 + 27x^2 + 1200 = 0$$

campbell_st · Answer

ok... that makes a huge difference

divide every term by - 3 and you get

$$x^4 - 9x^2 - 400 = 0$$

this factors to 

$$(x^2 - 25)(x^2 + 16) = 0$$

now you can go and find all the solutions...

anonymous · Answer

So the equation might actually come out as x = 4, 5i?

campbell_st · Answer

well they won't be the solutions.... you need to solve

$$x^2 - 25 = 0~~~~~~ and~~~~~~x^2 + 16 = 0$$

and there are complex solutions.

anonymous · Answer

Wait, I think something's wrong with my math. For the quadratic equation, I ended up with:$$x^2 = \frac{ -27 - 123 }{ 6 } , \frac{ -27+123 }{ 6 }$$

anonymous · Answer

But the answers should be 4i and 5, not 5i and 4?

anonymous · Answer

Can someone help me figure out what I did wrong?

campbell_st · Answer

wow.... ok... you need to work through the problem and not work to the solution... 

the equation is degree 4 so you need to show 4 solutions.... 


$$x^2 - 25 = 0$$

add 25 to both sides of the equation 

$$x^2 = 25$$

solve this

and $$x^2 + 16 = 0$$

subtract 16 from both sides of the equation 

$$x^2 = -16$$

now solve for x

anonymous · Answer

x = 4i, and x also equals 5.

campbell_st · Answer

well they are only 2 of the 4 solutions....

campbell_st · Answer

$$x^2 = 25~~~~~x = \pm 5$$

you can find the other.

anonymous · Answer

What about the negative forms of the answers, since they would be eliminated after being raised to the 4th and 2nd power?

campbell_st · Answer

the solutions are 

$$x = -5, 5 -4i, 4i$$

substitute any solution to see if it worked.. 

you can't have a degree 4 equation with 2 solutions...

anonymous · Answer

Thank you for your help!