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

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Answer 1 · 2015-04-30T12:20:19

The equation is of the form $Ax^2+Bx+C=0$ where:

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

Now the disciminant
$D=B^2-4*A*C=$
$(-6)^2-4*9*1=$
$36-36=0$

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:

$x=(-B+sqrtD)/(2*A) or x=(-B-sqrtD)/(2*A)$

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

$x=(-B)/(2*A)=(-(-6))/(2*9)=1/3$

This means the graph (a 'valley'-parabola) will just touch the $x$ -axis at the point $(1/3,0)$

How do you find the discriminant and how many solutions does 9x^2-6x+1=09x^2-6x+1=0 have?