x is a number such that x2+3x+9=0. What is the value of x3?
since (x-3)(x^2+3x+9) = x^3 - 3^3
x^3 - 3^3 = 0
x^3 = 27
What a great way to do this question.
Did you see the round-about way I used ??
http://www.jiskha.com/display.cgi?id=1360597831
I was even thinking of using DeMoivre's theorem.
I was going to show an alternative direct solution, but when I saw where it was headed, I blew it off.
Can't believe you
(a) slogged through it all
(b) got it right!
To solve this problem, we need to find the value of x in the equation x^2 + 3x + 9 = 0 and then calculate x^3.
Step 1: Solve the quadratic equation x^2 + 3x + 9 = 0 to find the value of x.
We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.
In our case, a = 1, b = 3, and c = 9. Substituting these values into the quadratic formula, we get:
x = ( -3 ± √(3^2 - 4*1*9)) / (2*1)
x = ( -3 ± √(9 - 36)) / 2
x = ( -3 ± √(-27)) / 2
The square root of a negative number is an imaginary number, so we have complex solutions:
x = ( -3 ± √(27)i) / 2
x = ( -3 ± 3√3i) / 2
Step 2: Calculate x^3 using the obtained value of x.
Using the properties of complex numbers, we can calculate x^3. Let's take x = ( -3 + 3√3i) / 2 for simplicity.
x^3 = [( -3 + 3√3i) / 2]^3
Using the formula (a + bi)^3 = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3, we can expand x^3 as follows:
x^3 = ( -3)^3 + 3( -3)^2(3√3i) + 3( -3)(3√3i)^2 + (3√3i)^3
x^3 = -27 + 3(9)(3√3i) + 3( -3)(-9i) + (27√3i^2)
x^3 = -27 + 81√3i - 81i + (27√3)(-1)
x^3 = -27 + 81√3i - 81i - 27√3
x^3 = -54 + 54√3i
So, the value of x^3 is -54 + 54√3i.
Please note that this answer may vary depending on the value of x you choose from the complex solutions.