Question

1. the two parabolas, $y = 2x^{2}-sqrt{3}x + 3$ and $y=x^{2}+sqrt{3}x + 6$, are graphed below. show that the parabolas intersect at $x=sqrt{6}+sqrt{3}$.

Explanation:

Step1: Set the two equations equal

Set $2x^{2}-\sqrt{3}x + 3=x^{2}+\sqrt{3}x + 6$.

Step2: Rearrange the equation

Move all terms to one - side: $2x^{2}-x^{2}-\sqrt{3}x-\sqrt{3}x+3 - 6=0$, which simplifies to $x^{2}-2\sqrt{3}x - 3=0$.

Step3: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-2\sqrt{3}$, and $c=-3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-2\sqrt{3})^{2}-4\times1\times(-3)=12 + 12=24$. Then $x=\frac{2\sqrt{3}\pm\sqrt{24}}{2}=\frac{2\sqrt{3}\pm2\sqrt{6}}{2}=\sqrt{3}\pm\sqrt{6}$.
We take the positive root $x=\sqrt{6}+\sqrt{3}$ since the context of the intersection point (usually considering real - valued and relevant solutions in the graph context) and when we substitute $x=\sqrt{6}+\sqrt{3}$ into the original equations, it satisfies the condition of intersection.

Answer:

We have shown through setting the equations equal, rearranging to a quadratic form, and using the quadratic formula that one of the intersection points of the two parabolas is $x=\sqrt{6}+\sqrt{3}$.