Question

The sum of two consecutive positive integers and their product is \[271\] .What are the integers?

Answer 1

Hint: In order to solve this question, first we will let the two consecutive positive integers. Then according to the given condition, we will make the equation and simplify the equation. Thus, we will get a quadratic equation. Later we will solve the quadratic equation. Hence, we get the required result. And remember that they both are positive integers, and they should be consecutive.

Complete step by step answer:
We have given that the sum of two consecutive positive integers and their product is \[271\].So, let us assume the first positive integer is \[x\] then the second positive integer will be \[x + 1\]. Now according to the question, we have
\[x + \left( {x + 1} \right) + x\left( {x + 1} \right) = 271\]
On multiplying, we get
\[ \Rightarrow x + \left( {x + 1} \right) + \left( {{x^2} + x} \right) = 271\]
Combining the like terms, we get
\[ \Rightarrow {x^2} + 3x + 1 = 271\]
Subtracting \[1\] from both the sides of the equation, we get
\[ \Rightarrow {x^2} + 3x = 270\]

Subtracting \[270\] from both the sides of the equation, we get
\[ \Rightarrow {x^2} + 3x - 270 = 0\]
Now we know that
The general form of a quadratic equation is \[a{x^2} + bx + c = 0\] and the value of \[x\] is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here, \[a = 1,{\text{ }}b = 3,{\text{ }}c = - 270\]
Therefore, we get
\[x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 1 \right)\left( { - 270} \right)} }}{{2\left( 1 \right)}}\]
Simplifying the above, we get
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 + 1080} }}{2}\]
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {1089} }}{2}\]

As we know that
\[\sqrt {1089} = 33\]
Therefore, we get
\[ \Rightarrow x = \dfrac{{ - 3 \pm 33}}{2}\]
Thus, we get two values as,
\[x = \dfrac{{ - 3 + 33}}{2}\] and \[x = \dfrac{{ - 3 - 33}}{2}\]
Solving the above values, we get
\[x = \dfrac{{30}}{2} = 15\] and \[x = \dfrac{{ - 36}}{2} = - 18\]
Since, it is given that the two numbers are positive integers.
Therefore, we get \[x = 15\]
And the second number i.e., \[x + 1 = 15 + 1 = 16\]
Therefore, two numbers are \[15\] and \[16\].

Hence, the two consecutive positive integers are \[15\] and \[16\].

Note: We can even take \[x\] and \[x - 1\] as both consecutive numbers and make an equation and solve for \[x\] . In this case we will get \[16\] as \[x\] and \[15\] as \[x - 1\]. Therefore, we have the same set of numbers. You can try it out yourself in the same procedure. Also be careful while doing the calculation part to avoid making errors.