Question

How do you find two consecutive integers whose product is $58$.

Answer 1

Hint: From the question given we have to find the two consecutive integers whose product is $58$. To solve the question first we have to assume the smallest number of the consecutive integers as ‘n’ then the other number will be $n+1$ because these two numbers are consecutive integers. For the number given in the question that is $58$ can not be written as the two consecutive integers.

Complete step by step answer:
From the question we have to find the two consecutive integers whose product is $58$.
Let us assume the one of the consecutive integers is $'n'$ then automatically the another integer will be $n+1$.
Therefore, the assumed consecutive integers are
$\Rightarrow n\,and\,n+1$
From the question given the product of these two numbers is $58$.
By this we will get the algebraic equation or quadratic equation.
The quadratic equation is
$\Rightarrow \left( n \right)\left( n+1 \right)=58$
By further simplifying we will get
$\Rightarrow {{n}^{2}}+n=58$
$\Rightarrow {{n}^{2}}+n-58=0$
As we know that the formula for the obtaining the roots of the quadratic equation$a{{x}^{2}}+bx+c=0$ given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ then
Then by comparing the above equation substitute the values in their respective positions,
By substituting the values in their respective positions, we will get,
$\Rightarrow n=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 1\times -58}}{2}$
$\Rightarrow n=\dfrac{-1\pm \sqrt{1+232}}{2}$
$\Rightarrow n=\dfrac{-1\pm \sqrt{233}}{2}$
Therefore, as we do not have a whole number for square root of $233$.

Hence, we do not have such a pair of consecutive integers whose product is $58$. Hence no solutions.

Note: Students can solve this question by writing factors of $58$ because product of a number means they are factors of the number. For suppose the factors of $58$ are $\left\{ 1,2,29,58 \right\}$ in this there are no two consecutive integers whose product is $58$ hence no solutions.