Question

16. The product of two consecutive negative integers is 10506. Write a quadratic equation that you could solve to find the integers.

Answer 1

Let's call x a negative number. We want the product of x and the one next number to it (x + 1) to be 10506. Then, we can write:

$x(x+1)=10506$

Now, we can apply distributive property on the parentheses, and rest 10506 on both sides:

$x^2+x-10506=0$

And now, we have a quadratic equation in standard form. We can solve this using the quadratic formula:

$\begin{gathered} x_{1,2}=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-10506)}}{2\cdot1}=\frac{-1\pm\sqrt{42025}}{2}=\frac{-1\pm205}{2} \\ . \\ x_1=\frac{-1+205}{2}=\frac{204}{2}=102 \\ . \\ x_2=\frac{-1-205}{2}=\frac{-206}{2}=-103 \end{gathered}$

Since we want x to be a negative answer, we take the negative solution, x = -103, And now, we can find the other number:

x + 1 = -103 + 1 = -102

Now, we can verify that the pair of numbers that we have found really are a solution for what we're looking for:

$(-103)(-102)=103\cdot102=10506$

Thus, the equation to solve this is:

$x(x+1)=10506$

And the solution is:

x = -103

x + 1 = -102