9xsquared - 6x + 1 = 0

first use discriminate to determine whether the equation has two nonreal complex solution, one solution with a multiplicity of two, or two real solutions, and then solve the equation.

Can someone put me in the right direction me this? thanks

2. Originally Posted by likewhoa
9xsquared - 6x + 1 = 0

first use discriminate to determine whether the equation has two nonreal complex solution, one solution with a multiplicity of two, or two real solutions, and then solve the equation.

Can someone put me in the right direction me this? thanks
What in particular are you having trouble with? The question is very clear...

Here you have a Quadratic Equation:

$\displaystyle 9x^2 - 6x + 1 = 0$ so $\displaystyle a = 9, b = -6, c = 1$.

Discriminant:

$\displaystyle \Delta = b^2 - 4ac$.

If $\displaystyle \Delta < 0$ the solutions are complex conjugates

If $\displaystyle \Delta = 0$ you have one repeated solution

If $\displaystyle \Delta > 0$ you have two real solutions.

The roots:

$\displaystyle x = \frac{-b \pm \sqrt{\Delta}}{2a}$.