Question

The product of two consecutive negative integers is 1122. How do you find the integers?

Answer 1

We first assume the negative integers and use the condition of the product of two consecutive negative integers being 1122. We get a quadratic equation of $n$. We use the quadratic formula to solve the value of the $n$. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete step-by-step solution:
We have been given that the product of two consecutive negative integers is 1122.
Let’s assume the smaller number is $n$. The number is a negative integer.
The other number will be $\left( n+1 \right)$. The numbers are consecutive.
So, the product of $n$ and $\left( n+1 \right)$ is 1122. This gives $n\left( n+1 \right)=1122$.
The equation becomes a quadratic equation ${{n}^{2}}+n-1122=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have ${{n}^{2}}+n-1122=0$. The values of a, b, c is $1,1,-1122$ respectively.
We put the values and get $n$ as $n=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times \left( -1122 \right)\times 1}}{2\times 1}=\dfrac{-1\pm \sqrt{4489}}{2}=\dfrac{-1\pm 67}{2}=-34,33$
The roots of the equation are real numbers.
So, values of $n$ are $n=-34,33$.
Now $n$ has to be negative. So, $n=-34$. The other number is $n+1=-34+1=-33$.
The integers are $-34,-33$.

Note: We can also use a grouping method. We break $n$ in ${{n}^{2}}+n-1122=0$.
$\begin{aligned} & {{n}^{2}}+n-1122=0 \\\ & \Rightarrow {{n}^{2}}+34n-33n-1122=0 \\\ & \Rightarrow \left( n+34 \right)\left( n-33 \right)=0 \\\ \end{aligned}$
The two solutions are $n=-34,33$ which gives the negative integers as $n=-34$.
The integers are $-34,-33$.