Question

what are the roots of the quadratic function?
y = 3r² - 2r - 6
a r = -1 and r = 3/4
b r = 1 and r = -3/4
c r = 1 and r = 3/4
d r = 1 and r = -4/3
e r = 2 and r = -3

Explanation:

Step1: Factor the quadratic equation

We have the quadratic function \( y = 3r^{2}-2r - 1\). To find the roots, we set \( y = 0\), so we need to solve \( 3r^{2}-2r - 1=0\). We can factor this quadratic equation. We need two numbers that multiply to \( 3\times(- 1)=-3\) and add up to \(-2\). The numbers are \(-3\) and \(1\). So we can rewrite the middle term:
\(3r^{2}-3r + r-1 = 0\)
Factor by grouping:
\(3r(r - 1)+1(r - 1)=0\)
\((3r + 1)(r - 1)=0\)

Step2: Solve for r

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).

• Case 1: \(r - 1=0\), then \(r=1\).
• Case 2: \(3r + 1=0\), then \(3r=-1\), and \(r=-\frac{1}{3}\).

Answer:

D. \(r = 1\) and \(r=-\frac{1}{3}\)