The roots of a quadratic equation ax²+bx+c=0 can be determined by calculating the discriminant Δ=b²-4ac.
If Δ>0, there are two distinct real solutions.
If Δ=0, there is one real solution.
If Δ<0, there are no real solutions and two complex solutions.
\(3x^2-6x+3=0 \\ \\
a=3 \\ b=-6 \\ c=3 \\ \\
\Delta=b^2-4ac=(-6)^2-4 \times 3 \times 3=36-36=0\)
Δ=0, so there is one real solution. The answer is A.
The solution is:
\(x=\frac{-b}{2a}=\frac{-(-6)}{2 \times 3}=\frac{6}{6}=1\)
