To determine the values of [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] needed to write the given quadratic equation in standard form, let's start with the given equation:
[tex]\[
\frac{2}{3}(x-4)(x+5) = 1
\][/tex]
First, we need to expand the left-hand side expression, [tex]\((x-4)(x+5)\)[/tex]:
1. Expand the binomials:
[tex]\[
(x-4)(x+5) = x^2 + 5x - 4x - 20 = x^2 + x - 20
\][/tex]
2. Next, multiply each term by [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[
\frac{2}{3} \cdot (x^2 + x - 20) = \frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3}
\][/tex]
Now, substitute this expanded form back into the equation:
[tex]\[
\frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3} = 1
\][/tex]
To express this in the standard quadratic form [tex]\(Ax^2 + Bx + C = 0\)[/tex], we need to set the right-hand side to zero by subtracting 1 from both sides:
[tex]\[
\frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3} - 1 = 0
\][/tex]
Simplify the right side (note that [tex]\(-1\)[/tex] is equivalent to [tex]\(-\frac{3}{3}\)[/tex] in fractional form):
[tex]\[
\frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3} - \frac{3}{3} = 0
\][/tex]
Combine the constants:
[tex]\[
\frac{2}{3}x^2 + \frac{2}{3}x - \frac{43}{3} = 0
\][/tex]
Comparing this with the standard form [tex]\(Ax^2 + Bx + C = 0\)[/tex], we identify the coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] as follows:
[tex]\[
A = \frac{2}{3}, \quad B = \frac{2}{3}, \quad C = -\frac{43}{3}
\][/tex]
The numerical results verify that:
[tex]\[
A = 0.6666666666666666, \quad B = 0.6666666666666666, \quad C = -14.333333333333334
\][/tex]
Therefore, the correct values for the coefficients are:
\[
A = \frac{2}{3}, \quad B = \frac{2}{3}, \quad C = -\frac{43}{3}
