Question

Find two consecutive natural numbers whose product is 20.

Answer 1

$\bf{ \red{step \: 1 - }decide \: the \: numbers}$

Let the two consecutive natural numbers be x and x+1 .

A/Q

$\bf{ \green{step \: 2 - }multiply \: them}$

(x)(x+1) = 20

$\bf{ \blue{step \: 3 - }solve \: quadrati c \: equation}$

x^2 + x =20

${x}^{2} + x - 20 = 0$

${x}^{2} + (5 - 4)x - 20 = 0$

${x}^{2} + 5x - 4x - 20 = 0$

$x(x + 5) - 4(x + 5) = 0$

$(x - 4)(x + 5) = 0$

$x - 4 = 0$

____________________________

$\huge{\underline{\bf{Case\:1}}}$

x=4

x+1 = 4+1 = 5

_____________________________

$\huge{\underline{\bf{Case\:2}}}$

x+5=0

x=-5

x+1 = -5+1

=-4

______________________________

$\textbf{So the two numbers can either} \\ \textbf{be 4 and 5 or -5 and -4}$

But question demands natural numbers.

Natural Numbers starts from 1 .

So, the two numbers are :- 4 and 5.


$\huge{\bf{Hope\:It\:Helps\: :)}}$

Answer 2

step1−decidethenumbers 

Let the two consecutive natural numbers be x and x+1 . 

A/Q 

\bf{ \green{step \: 2 - }multiply \: them}step2−multiplythem 

(x)(x+1) = 20 

\bf{ \blue{step \: 3 - }solve \: quadrati c \: equation}step3−solvequadraticequation 

x^2 + x =20 

{x}^{2} + x - 20 = 0x2+x−20=0 

{x}^{2} + (5 - 4)x - 20 = 0x2+(5−4)x−20=0 

{x}^{2} + 5x - 4x - 20 = 0x2+5x−4x−20=0 

x(x + 5) - 4(x + 5) = 0x(x+5)−4(x+5)=0 

(x - 4)(x + 5) = 0(x−4)(x+5)=0 

x - 4 = 0x−4=0 

____________________________

\huge{\underline{\bf{Case\:1}}}Case1​ 

x=4 

x+1 = 4+1 = 5 

_____________________________

\huge{\underline{\bf{Case\:2}}}Case2​ 

x+5=0 

x=-5 

x+1 = -5+1 

=-4 

______________________________

\begin{lgathered}\textbf{So the two numbers can either} \\ \textbf{be 4 and 5 or -5 and -4}\end{lgathered}So the two numbers can eitherbe 4 and 5 or -5 and -4​ 

But question demands natural numbers. 

Natural Numbers starts from 1 . 

So, the two numbers are :- 4 and 5.