To solve this problem, let's define the two consecutive odd integers.

Let:

• $x$ be the greater integer.
• The smaller integer would then be $x - 2$ (since consecutive odd integers differ by 2).

Their product is given as 143, so the equation representing this scenario is:

$x \cdot (x - 2) = 143$

Now, let's expand and solve for $x$:

$x^2 - 2x = 143$

Rearranging the equation to standard quadratic form:

$x^2 - 2x - 143 = 0$

To solve for $x$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -2$, and $c = -143$. Plugging in these values:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-143)}}{2 \cdot 1}$

$x = \frac{2 \pm \sqrt{4 + 572}}{2}$

$x = \frac{2 \pm \sqrt{576}}{2}$

Since $\sqrt{576} = 24$, we have:

$x = \frac{2 \pm 24}{2}$

This gives two solutions:

1. $x = \frac{2 + 24}{2} = 13$
2. $x = \frac{2 - 24}{2} = -11$

Since we are looking for positive integers, $x = 13$.

Thus, the greater integer is:

$\boxed{13}$

Equation Completion

The equation provided in the question is:

$x(x - \, \underline{2} \,) = 143$

Here, the blank should be filled with 2.

Would you like more details or have any questions?

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Related Questions:

1. What are the steps to solve a quadratic equation using the quadratic formula?
2. How can you determine if a quadratic equation has real or imaginary solutions?
3. Can you find another set of consecutive integers whose product is a perfect square?
4. How do you solve for consecutive even integers if their product is given?
5. What happens if the product of two integers is negative? How do you solve for them?

Tip:

When solving for consecutive odd or even integers, remember they always differ by 2 units.