Answer:
x = { (-1+√2i)/3 and (-1 -√2i)/3)}
Explanation:
The given equation is a quadratic equation.
9x^2 + 6x + 3 = 0
Here 3 is the common factor.
3(3x^2 + 2x + 1) = 0
Dividing both sides by 3, we get
3x^2 + 2x + 1 = 0
Here the value of a = 3, b = 2 and c = 1 by comparing the given equation with the general form of quadratic equation y = ax^2 + bx + x
We have quadratic formula to find the solution.
x = (-b ± √(b^2 - 4ac))/ 2a
Now we have to plug in a = 3, b = 2 and c =1 in the above formula, we get
x = (-2 ± √(2^2 - 4*3*1)) / 2*3
x = (-2±√-8) /6
√-8 = √-1√4√2 = 2√2i [√-1 = i]
Therefore, x = (-2±2√2i) / 6
Here the roots are complex.
we can take out 2 in the numerator because it is a common factor.
x = 2(-1 ±√2i)/6
Simplifying the above, we get
x = (-1 ±√2i)/3
There are two roots
x = { (-1+√2i)/3 and (-1 -√2i)/3)}
Hope you will understand the concept.
Thank you.