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The solutions to the equation are \( x = -6 \) and \( x = \frac{2}{3} \).
Solución
We start by checking the two candidate pairs of solutions. When we have a quadratic with two roots \( r_1 \) and \( r_2 \), we know that (up to a nonzero constant multiple) the equation can be written in factored form as
\[
a(x - r_1)(x - r_2)=0.
\]
Two things to verify are the sum and product of the roots.
If the two roots are \( r_1 \) and \( r_2 \), then by Vieta’s formulas for a quadratic \( ax^2+bx+c=0 \) (with \(a\ne 0\)) we have:
\[
r_1 + r_2 = -\frac{b}{a} \quad \text{and} \quad r_1r_2 = \frac{c}{a}.
\]
Let’s compare the candidate answer options:
1. If the proposed solutions are
\[
x = -6 \quad \text{and} \quad x = -\frac{2}{3},
\]
then
\[
r_1 + r_2 = -6 - \frac{2}{3} = -\frac{20}{3},\quad \text{and} \quad r_1r_2 = \left(-6\right)\left(-\frac{2}{3}\right) = 4.
\]
2. If the proposed solutions are
\[
x = -6 \quad \text{and} \quad x = \frac{2}{3},
\]
then
\[
r_1 + r_2 = -6 + \frac{2}{3} = -\frac{16}{3},\quad \text{and} \quad r_1r_2 = \left(-6\right)\left(\frac{2}{3}\right) = -4.
\]
3. If the proposed solutions are
\[
x = 6 \quad \text{and} \quad x = \frac{2}{3},
\]
then
\[
r_1 + r_2 = 6 + \frac{2}{3} = \frac{20}{3},\quad \text{and} \quad r_1r_2 = 6\left(\frac{2}{3}\right) = 4.
\]
Without the original equation explicitly given, one common scenario is that the quadratic’s coefficients are such that the factored form (after clearing fractions) is
\[
3(x + 6)\Bigl(x - \frac{2}{3}\Bigr) = 0.
\]
Expanding this gives:
\[
3\left[x^2 - \frac{2}{3}x + 6x - 4\right] = 3\left[x^2 + \frac{16}{3}x - 4 \right] = 3x^2 + 16x -12 = 0.
\]
Here the sum and product of the roots are:
\[
r_1 + r_2 = -\frac{16}{3} \quad \text{and} \quad r_1r_2 = \frac{-12}{3} = -4,
\]
which exactly match the values from option 2.
Thus, the solutions to the equation are
\[
x = -6 \quad \text{and} \quad x = \frac{2}{3}.
\]
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