Given the equation in standard form, provide the values of [tex]$a$[/tex], [tex]$b$[/tex], and [tex]$c$[/tex]:

[tex]\[
\frac{2}{3}(x-4)(x+5)=1
\][/tex]

A. [tex]$a = 2; b = 2; c = -43$[/tex]

B. [tex]$a = \frac{2}{3}; b = 1; c = -20$[/tex]

C. [tex]$a = 2; b = 2; c = 43$[/tex]



Answer :

To convert the given equation

[tex]$ \frac{2}{3}(x-4)(x+5) = 1 $[/tex]

into its standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], we will follow these steps:

1. Expand the left-hand side: Start by expanding the left side of the equation.

[tex]\[ \frac{2}{3}(x-4)(x+5) \][/tex]

Use the distributive property (FOIL method) to expand [tex]\( (x-4)(x+5) \)[/tex]:

[tex]\[ (x-4)(x+5) = x^2 + 5x - 4x - 20 = x^2 + x - 20 \][/tex]

Now multiply this result by [tex]\( \frac{2}{3} \)[/tex]:

[tex]\[ \frac{2}{3} (x^2 + x - 20) = \frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3} \][/tex]

2. Subtract 1 from both sides: To move everything to the left-hand side and set the equation to zero, subtract 1 from both sides of the equation:

[tex]\[ \frac{2}{3}x^2 + \frac{2}{3}x - \frac{40}{3} - 1 = 0 \][/tex]

3. Combine constants: [tex]\( - \frac{40}{3} - 1 \)[/tex] can be written as [tex]\( - \frac{40}{3} - \frac{3}{3} = - \frac{43}{3} \)[/tex].

So the equation becomes:

[tex]\[ \frac{2}{3}x^2 + \frac{2}{3}x - \frac{43}{3} = 0 \][/tex]

Now we have the equation in standard form:

[tex]\[ a = \frac{2}{3}, \quad b = \frac{2}{3}, \quad c = - \frac{43}{3} \][/tex]

Given the options:

[tex]\[ (A)\; 2 \quad B = 2 \quad C = -43 \\ (B)\; \frac{2}{3} \quad B = 1 \quad C = -20 \\ (C)\; 2 \quad B = 2 \quad C = 43 \][/tex]

None of the provided options match exactly with the calculated coefficients [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex]. However, our correct values are:

[tex]\[ a = \frac{2}{3}, \quad b = \frac{2}{3}, \quad c = - \frac{43}{3} \][/tex]

Therefore, if forced to choose from given options, consult the problem constraints and decide accordingly. However, our mathematically correct coefficients do not align with any of the provided options verbatim.