Question

Find two consecutive natural numbers whose product is 20.

Answer 1

Hint: When the preceding number is one more than the original number then the two numbers are known as consecutive numbers.
In this question we will consider two consecutive natural numbers and will obtain a quadratic equation by multiplying them and by solving the equation we will find the natural numbers which will give 20 as their product.

Complete step-by-step answer:
Given
Two consecutive numbers whose product is 20
Let the two consecutive number whose product is 20 be \[n\] and \[n + 1\]
Now since product of two consecutive natural numbers is 20, hence we can write this as
 \[n \times \left( {n + 1} \right) = 20\]
By solving the obtained equation, we get
 \[
  {n^2} + n = 20 \\
   \Rightarrow {n^2} + n - 20 = 0 - - (i) \;
 \]
Now since we have obtained the equation in quadratic form so we will find the two roots of the equation to find the numbers by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4\left( a \right)\left( c \right)} }}{{2a}}\] , hence we can write
 \[
  n = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 20} \right)} }}{{2 \times 1}} \\
   = \dfrac{{ - 1 \pm \sqrt {1 + 80} }}{2} \\
   = \dfrac{{ - 1 \pm \sqrt {81} }}{2} \\
   = \dfrac{{ - 1 \pm 9}}{2} \;
 \]
This can be written as
 \[
  n = \dfrac{{ - 1 + 9}}{2},\dfrac{{ - 1 - 9}}{2} \\
   = \dfrac{8}{2},\dfrac{{ - 10}}{2} \\
   = 4, - 5 \;
 \]
Hence we get the roots of the quadratic equation \[n = 4, - 5\]
Now since the two consecutive naturals number are positive so \[n = - 5\] cannot be consider, hence we can say one of the consecutive number is \[n = 4\]
Therefore the other number will be \[n + 1 = 4 + 1 = 5\]
Hence we can say two consecutive natural numbers whose product is 20 are \[4\& 5\] .
To verify:
 \[
  n \times \left( {n + 1} \right) = 20 \\
   \Rightarrow 4 \times 5 = 20 \\
   \Rightarrow 20 = 20\left( {Verified} \right) \;
 \]
So, the correct answer is “ \[4\& 5\] ”.

Note: Another method to find the roots of a quadratic equation is by splitting the middle term of the quadratic equation.
In this question the roots of the obtained quadratic equation can also be found by splitting \[ - n\] in such a way that the product of the splinted terms gives the third term of their equation and their sum is equal to the middle term.
  \[
  {n^2} + n - 20 = 0 \\
   \Rightarrow {n^2} + 5n - 4n - 20 = 0 \\
   \Rightarrow n\left( {n + 5} \right) - 4\left( {n + 5} \right) = 0 \\
   \Rightarrow \left( {n - 4} \right)\left( {n + 5} \right) = 0 \;
 \]