Hi,
I have a problem solving the following:
27x5+8x2=0
Help would be appreciated :)
Solve for x:
27 x^5+8 x^2 = 0
The left hand side factors into a product with three terms:
x^2 (3 x+2) (9 x^2-6 x+4) = 0
Split into three equations:
x^2 = 0 or 3 x+2 = 0 or 9 x^2-6 x+4 = 0
Take the square root of both sides:
x = 0 or 3 x+2 = 0 or 9 x^2-6 x+4 = 0
Subtract 2 from both sides:
x = 0 or 3 x = -2 or 9 x^2-6 x+4 = 0
Divide both sides by 3:
x = 0 or x = -2/3 or 9 x^2-6 x+4 = 0
Divide both sides by 9:
x = 0 or x = -2/3 or x^2-(2 x)/3+4/9 = 0
Subtract 4/9 from both sides:
x = 0 or x = -2/3 or x^2-(2 x)/3 = -4/9
Add 1/9 to both sides:
x = 0 or x = -2/3 or x^2-(2 x)/3+1/9 = -1/3
Write the left hand side as a square:
x = 0 or x = -2/3 or (x-1/3)^2 = -1/3
Take the square root of both sides:
x = 0 or x = -2/3 or x-1/3 = i/sqrt(3) or x-1/3 = -i/sqrt(3)
Add 1/3 to both sides:
x = 0 or x = -2/3 or x = 1/3+i/sqrt(3) or x-1/3 = -i/sqrt(3)
Add 1/3 to both sides:
Answer: |x = 0 or x = -2/3 or x = 1/3+i/sqrt(3) or x = 1/3-i/sqrt(3)
+9665 \(\begin{array}{rll}27x^5+8x^2&=&0\\27x^3+8&=&0 \\(3x+2)(9x^2-6x+4)&=&0\\\\ 3x_1+2 = 0&&9(x_2)^2-6x_2+4=0\\ x_1=-\frac{2}{3}&&x_2=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(9)(4)}}{2(9)}\\ &&\;\;\;\;\;\!\;\;\!\!=\dfrac{6\pm6i\sqrt{3}}{18}\\ &&\;\;\;\;\;\!\;\;\!\!=\dfrac{1\pm \sqrt{3}i}{3}\end{array}\\ x_1 = -\frac{2}{3}\\x_2 = \frac{1}{3}+\frac{i}{\sqrt3}\\x_3=\frac{1}{3}-\frac{i}{\sqrt3}\)
Well was I doing something wrong or I have just found the complex solutions for you?
