# How do you find the discriminant and how many and what type of solutions does 4x^2-2x+1=0 have?

May 5, 2015

The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 4 , b = - 2 , c = 1$

The Disciminant is given by :
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(- 2\right)}^{2} - \left(4 \cdot 4 \cdot 1\right)$
$= 4 - 16 = - 12$

If $\Delta = 0$ then there is only one solution.
(for $\Delta > 0$ there are two solutions,
for $\Delta < 0$ there are no real solutions)

As $\Delta = - 12$, this equation has NO REAL SOLUTIONS

• Note :
The solutions are normally found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

Substituting the value of $\Delta$ will give us 2 imaginary roots/solutions.