At how many points does the graph of a quadratic equation cut x axis

Quadratic equation is of the form \(y=ax^2+bx+c\). at \(x\) intercept \(y=0\)

hence \(ax^2+bx+c=0 => x^2+\frac{b}{a}x+\frac{c}{a}=0\)

we need to find the value of \(\frac{b}{a}\) & \(\frac{c}{a}\) to arrive at the equation and hence at the \(x\)-intercepts

Statement 1: From Vieta's formula , sum of roots\(=-\frac{b}{a}=5\), hence \(\frac{b}{a}=-5\) but nothing mentioned about \(\frac{c}{a}\). Insufficient

Statement 2: Again from Vieta's formulas, product of roots\(=\frac{c}{a}=6\). but nothing mentioned about \(\frac{b}{a}\). Insufficient

Combining 1 & 2 we get all the required values, hence our equation is \(x^2-5x+6=0\)

Find the roots of this equation to know the \(x\)-intercepts

\(x^2-5x+6=(x-3)(x-2)=0\). therefore \(x\)-intercepts are \(3\) and \(2\). Sufficient

Option C