Which two values of x are roots of the polynomial below? (select all that apply
3x^2-3x+1

x = (3 + √-3) / 6 and x = (3 - √-3) / 6, so answers A and B are correct.

Step-by-step explanation:

To clarify, when we say "roots" of a quadratic expression, we're talking about it's x-intercepts, or where the entire expression = 0.  Let's start with that:

3x² - 3x + 1 = 0

Let's complete the square to solve it:

9x² - 9x + 3 = 0

9x² - 9x + 2.25 = -0.75

(3x - 1.5)² = -0.75

(3x - 3 / 2)² = -3 / 4

3x - 3 / 2 = ± i√(3 / 4)

3x - 3 / 2 = ± i√3 / 2

3x = 3 / 2 ± i√3 / 2

3x = (3 ± i√3) / 2

x = (3 ± i√3) / 6

So the x is equal to both (3 + i√3) / 6 and (3 - i√3) / 6.

You'll note that these answers don't exactly match the given ones.  The only difference there is that I factored -1 out of the radical, giving you i as a coefficient to it. You may not yet be at a point where you've covered imaginary numbers, but I assure you the answers I found here are exactly the same as x = (3 + √-3) / 6 and x = (3 - √-3) / 6.

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point