
The product of two consecutive negative integers is 1122. How do you find the integers?
Answer
492.6k+ views
Hint: 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 ${{n}^{2}}+n-1122=0$.
$\begin{align}
& {{n}^{2}}+n-1122=0 \\
& \Rightarrow {{n}^{2}}+34n-33n-1122=0 \\
& \Rightarrow \left( n+34 \right)\left( n-33 \right)=0 \\
\end{align}$
The two solutions are \[n=-34,33\] which gives the negative integers as \[n=-34\].
The integers are \[-34,-33\].
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 ${{n}^{2}}+n-1122=0$.
$\begin{align}
& {{n}^{2}}+n-1122=0 \\
& \Rightarrow {{n}^{2}}+34n-33n-1122=0 \\
& \Rightarrow \left( n+34 \right)\left( n-33 \right)=0 \\
\end{align}$
The two solutions are \[n=-34,33\] which gives the negative integers as \[n=-34\].
The integers are \[-34,-33\].
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