Medal and all that Junk. What are real and complex solutions of this polynomial equation? 27x^3+125=0
sorry idk :(
\(27x^3+125 = (3x)^3 + (5)^3\), right? That is sum of cubes \(a^3 + b^3 = (a + b)(a^2 – ab + b^2) \)
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Okay, yes. I see that.
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OK, factoring given polynomial equation, you have \((3x + 5)(9x^2 + 15x + 25)=0\) So either \(3x+5=0\) or \(9x^2 + 15x + 25=0\) So you now have two equations to solve x for. For second equation, you can apply quadratic formula.
oops, \(9x^2-15x+25\)***
I see that!! Alright, I'll solve out, can you check?
Sure.
So the first is x=-5/3 right?
Yes.
Okay. Give me a second for the second.
I got it down to 15+- sqrt of -675/16
/18*****
Now I know I'll need to use an imaginary form of this, but how would i go about that?
Well, \(\sqrt{-675}=i\sqrt{675}\), right?
Yep.
hey i gave u the answer
I know. @amysparkly12 but I wanted to figure out how to do each step!!
And I am sure you need to simplify \(\sqrt{675}\) You can see that it is divisible by 25, so \(675= 25\times27 = 5^2\cdot3^3\) So \(i\sqrt{675} = 5\cdot3i\sqrt{3} = 15i\sqrt{3}\). Make sense so far?
Sort of, why don't I just find the sqrt of 675 at that point and divide by 18?
So you have \(\dfrac{15\pm15i\sqrt3}{18}\) You can factor 3 out and have them cancel: \(\dfrac{3(5\pm5i\sqrt3)}{3\cdot6}\) You can do that, but if whatever inside square root is not square number, you will get messy number. See: \(\sqrt{675}=25.98076211353316...\)
And we don't want messy number in our hands, right?
Ah, I see now. Okay so after you factor out 3??
Obviously, we then cancel 3 out. So we are left with \(x =\dfrac{5\pm5i\sqrt3}{6}\) So you have two complex numbers as answers: \(x=\dfrac{5+5i\sqrt3}{6}\) and \(x=\dfrac{5-5i\sqrt3}{6}\) Make sense so far?
Yes.
Now we just solve these two out?
Or are these our answers?
Well, you can arrange numbers around, but I think they are in their simplest form. You are done here.
Awesommmmeeeee. Gah. Okay, thank you so much!
No problem!
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