# How do you find the discriminant and how many solutions does 9x^2-6x+1=0 have?

Apr 30, 2015

The equation is of the form $A {x}^{2} + B x + C = 0$ where:

$A = 9 , B = - 6 , C = 1$

Now the disciminant
$D = {B}^{2} - 4 \cdot A \cdot C =$
${\left(- 6\right)}^{2} - 4 \cdot 9 \cdot 1 =$
$36 - 36 = 0$

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

Extra:
The solutions are normally found by working out:

$x = \frac{- B + \sqrt{D}}{2 \cdot A} \mathmr{and} x = \frac{- B - \sqrt{D}}{2 \cdot A}$

But since $D = 0$ in this case it comes down to:

$x = \frac{- B}{2 \cdot A} = \frac{- \left(- 6\right)}{2 \cdot 9} = \frac{1}{3}$

This means the graph (a 'valley'-parabola) will just touch the $x$-axis at the point $\left(\frac{1}{3} , 0\right)$