Answer :
To find a quadratic equation with roots [tex]\( \frac{5}{3} \)[/tex] and [tex]\( \frac{7}{3} \)[/tex], we can use the fact that if a quadratic equation has roots [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex], then it can be written in the form
[tex]$$
a\left(x^2 - (r_1 + r_2)x + r_1 r_2\right) = 0,
$$[/tex]
where [tex]\( a \)[/tex] is a nonzero constant. Here are the steps:
1. Compute the sum of the roots:
[tex]$$
r_1 + r_2 = \frac{5}{3} + \frac{7}{3} = \frac{5+7}{3} = \frac{12}{3} = 4.
$$[/tex]
2. Compute the product of the roots:
[tex]$$
r_1 \cdot r_2 = \frac{5}{3} \cdot \frac{7}{3} = \frac{35}{9}.
$$[/tex]
3. Without the multiplier [tex]\( a \)[/tex], the quadratic equation would look like this:
[tex]$$
x^2 - 4x + \frac{35}{9} = 0.
$$[/tex]
4. To eliminate the fraction and obtain integer coefficients, multiply the entire equation by [tex]\( 9 \)[/tex] (choosing [tex]\( a = 9 \)[/tex]):
[tex]$$
9 \left( x^2 - 4x + \frac{35}{9} \right) = 9x^2 - 36x + 35 = 0.
$$[/tex]
Thus, the quadratic equation in standard form is:
[tex]$$
9x^2 - 36x + 35 = 0.
$$[/tex]
Among the provided choices, this corresponds to option A.
[tex]$$
a\left(x^2 - (r_1 + r_2)x + r_1 r_2\right) = 0,
$$[/tex]
where [tex]\( a \)[/tex] is a nonzero constant. Here are the steps:
1. Compute the sum of the roots:
[tex]$$
r_1 + r_2 = \frac{5}{3} + \frac{7}{3} = \frac{5+7}{3} = \frac{12}{3} = 4.
$$[/tex]
2. Compute the product of the roots:
[tex]$$
r_1 \cdot r_2 = \frac{5}{3} \cdot \frac{7}{3} = \frac{35}{9}.
$$[/tex]
3. Without the multiplier [tex]\( a \)[/tex], the quadratic equation would look like this:
[tex]$$
x^2 - 4x + \frac{35}{9} = 0.
$$[/tex]
4. To eliminate the fraction and obtain integer coefficients, multiply the entire equation by [tex]\( 9 \)[/tex] (choosing [tex]\( a = 9 \)[/tex]):
[tex]$$
9 \left( x^2 - 4x + \frac{35}{9} \right) = 9x^2 - 36x + 35 = 0.
$$[/tex]
Thus, the quadratic equation in standard form is:
[tex]$$
9x^2 - 36x + 35 = 0.
$$[/tex]
Among the provided choices, this corresponds to option A.
