Find the values of two numbers whose sum is 27 and product is 182?

Hint: We start solving the problem by assigning the variables for the two given numbers. We use the condition that the sum of those two is 27 to get our first relation between them. We then use the condition that the product of those two numbers is 182 to get our second relation. We then solve these two relations to get the required values of two numbers.

Complete step by step answer:
According to the problem, we need to find the numbers whose sum is 27 and product is 182.
Let us assume the two required numbers are ‘a’ and ‘b’.
From the problem, we are given that the sum of these two numbers is 27.
So, we have $a+b=27$.
$\Rightarrow b=27-a$ ---(1).
From the problem, we are given that the product of these numbers is 182.
So, we have $ab=182$ ---(2).
Let us substitute equation (2) in equation (1).
So, we have $a\left( 27-a \right)=182$.
$\Rightarrow 27a-{{a}^{2}}=182$.
$\Rightarrow {{a}^{2}}-27a+182=0$.
$\Rightarrow {{a}^{2}}-13a-14a+182=0$.
$\Rightarrow a\left( a-13 \right)-14\left( a-13 \right)=0$.
$\Rightarrow \left( a-14 \right)\left( a-13 \right)=0$.
$\Rightarrow a-14=0$ or $a-13=0$.
$\Rightarrow a=14$ or $a=13$ ---(3).
Let us assume $a=14$ and substitute this in equation (1).
So, we have $b=27-14=13$.
∴ The two numbers are 14 and 13.
Let us assume $a=13$ and substitute this in equation (1).
So, we have $b=27-13=14$.

∴ The two numbers are 13 and 14.

Note: Whenever we get this type of problems, we first assign variables for the unknowns in the problem to avoid confusion and calculation mistakes. We can also solve the problem as shown below:
We have $a+b=27$ and $ab=182$.
Let us do squaring on both sides of $a+b=27$.
So, we get ${{\left( a+b \right)}^{2}}-4ab={{\left( a-b \right)}^{2}}$.
$\Rightarrow {{\left( 27 \right)}^{2}}-4\left( 182 \right)={{\left( a-b \right)}^{2}}$.
$\Rightarrow 729-728={{\left( a-b \right)}^{2}}$.
$\Rightarrow 1={{\left( a-b \right)}^{2}}$.
$\Rightarrow a-b=1$.
$\Rightarrow a-1=b$. Let us substitute this in $a+b=27$.
So, we have $a+a-1=27$.
$\Rightarrow 2a=28$.
$\Rightarrow a=14$. Let us substitute this in $a+b=27$.
So, we have $14+b=27\Rightarrow b=13$.
∴ The numbers are 14 and 13.