Question

Find two numbers whose sum is 27 and product is 182.

Answer 1

Hint: Assume x and y as the two numbers. Form first relation in x and y by taking their sum and equating with 27. Form a second relation in x and y by taking their product and equating with 182.
Solve the two equations by forming a quadratic equation and using the middle term split method. Obtain the value of x and get the answer.

Complete step-by-step answer:
Here, we have to find two numbers such that their sum is 27 and their product is 182.
Now, let us assume the two numbers as x and y. So, according to the question, we have,
\[\Rightarrow x+y=27\]
\[\Rightarrow y=27-x\] - (1)
And, \[xy=182\] - (2)
Substituting the value of y from equation (1) in equation (2), we get,
\[\begin{align}
  & \Rightarrow x\left( 27-x \right)=182 \\
 & \Rightarrow 27x-{{x}^{2}}=182 \\
 & \Rightarrow {{x}^{2}}-27x+182=0 \\
\end{align}\]
Splitting the middle term, we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-14x-13x+182=0 \\
 & \Rightarrow x\left( x-14 \right)-13\left( x-14 \right)=0 \\
\end{align}\]
Taking \[\left( x-14 \right)\] common we get,
\[\Rightarrow \left( x-14 \right)\left( x-13 \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow x-14=0\] or x – 13 = 0
\[\Rightarrow x=14\] or x = 13
1. When x = 14: -
\[\Rightarrow y=27-14=13\]
2. When x = 13: -
\[\Rightarrow y=27-13=14\]
Hence, there can be two possible ways to write the numbers whose sum is 27 and product is 182, they are (x, y) = (13, 14) or (x, y) = (14, 13). So, ultimately the two numbers are 13 and 14, which is our answer.

Note: One may note that we can also solve this question without using the second variable y. We can directly assume the two numbers as x and (27 – x) because we can clearly see that their sum will be 27. Now, note that we have solved the obtained quadratic equation using the middle term split method. One can also use the discriminant method given as: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to get the answer. Here, ‘a’ is the co – efficient of \[{{x}^{2}}\], ‘b’ is the co – efficient of x and ‘c’ is the constant term. As you can see that we have obtained two values of x after solving the quadratic equation and none of them is rejected. This is because both the values of x are valid.