Question 162093
9x^2+1=6x
First let's but this in standard form;
{{{9x^2-6x+1=0}}}
discriminant;
{{{b^2-4ac}}}
a=9
b=-6
c=1
Now we plug these in;
{{{-6^2-4(9)(1)}}}
36-36=0
Since the discriminant is 0, there is one real root.
Now use the discriminant in the quadratic equation to find the number;
{{{(-6+-sqrt(0))/2(9)}}}
{{{-6/18 = 1/3}}}
If the discriminant is negative, there are no real roots;
If the discriminant is positive, there are two real roots.
:)