0
let f(x) = 3x^2 - 3x + 1. The roots of f is the same as asking for x where f(x) = 0.
So we do it f(x) = 0, 3x^2 - 3x + 1 = 0. Using "complete the square" method
3(x^2 - x) + 1 = 0
3(x^2 + (2 . - 1/2 . x) + (-1/2)^2 -(-1/2)^2) + 1 = 0
3(x - 1/2)^2 -3/4 + 1 = 0
3(x - 1/2)^2 + 1/4 = 0
By the quadratic formula, the solution comes out to be
x = [3+√(-3)]/6 and x = [3-√(-3)]/6
or
x = (3+i√3)/6 and x = (3-i√3)/6
in other words, the solutions are complex numbers.
Still curious? Ask our experts.
Chat with our AI personalities
Blake
Devin
RafaAdd your answer:
Earn +20 pts
Which two values of x are roots of the polynomial x2 plus 5x plus 9?
-2.5 + 1.6583123951777i-2.5 - 1.6583123951777i
Which two values of x are roots of the polynomial below x2 plus 5x plus 11?
There are none because the discriminant of the given quadratic expression is less than zero.
X2-11x plus 13 has two values of x are roots of a polynomial what is it?
x=11+69/2 and x=11-69/2
Which two values of x are roots of the polynomial x2 plus 3x - 5?
To find the roots of the polynomial (x^2 + 3x - 5), we need to set the polynomial equal to zero and solve for x. So, (x^2 + 3x - 5 = 0). To solve this quadratic equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where a = 1, b = 3, and c = -5. Plugging these values into the formula, we get (x = \frac{-3 \pm \sqrt{3^2 - 41(-5)}}{2*1}), which simplifies to (x = \frac{-3 \pm \sqrt{29}}{2}). Therefore, the two values of x that are roots of the polynomial are (x = \frac{-3 + \sqrt{29}}{2}) and (x = \frac{-3 - \sqrt{29}}{2}).
What kind of polynomial is this 2x5 plus 2x4 plus 2x plus 1?
A fifth degree polynomial.
