Question

The average of a non-zero number and its square is \[5\] times the number. The number is
A.\[9\]
B.\[17\]
C.\[29\]
D.\[295\]

Answer 1

Hint: We will make an assumption about the number to be found. Average is the sum of all the observations divided by the number of the observations. Say, if we are told to find the average of n number of numbers, then the average will be the sum of all the numbers divided by the number n. We need to form a quadratic equation for this question.

Complete step-by-step answer:
We let the required number be \[x\].
The square of the number will be \[{x^2}\].
The average of the number and its square will be \[\dfrac{{x + {x^2}}}{2}\].
The question tells us that the average of the number and its square is equal to 5 times the number itself. Therefore, the average \[ = 5x\]
So, according to the question, we get the equation
\[
  \dfrac{{x + {x^2}}}{2} = 5x \\
   \Rightarrow x + {x^2} = 2 \times 5x \\
   \Rightarrow x + {x^2} = 10x \\
   \Rightarrow {x^2} = 10x - x \\
   \Rightarrow {x^2} = 9x \\
   \Rightarrow {x^2} - 9x = 0 \\
   \Rightarrow x(x - 9) = 0 \\
\]
Either \[x = 0\]or \[
  x - 9 = 0 \\
   \Rightarrow x = 9 \\
\]
The value of the number will not be 0 because it is specifically mentioned in the question that a non zero number is taken. Hence, \[x \ne 0\]
Therefore, the value of the number is \[9\], that is, \[x = 9\].

Thus, the answer is option A.

Note: We might make the mistake of dividing both the sides with x, but that will be wrong because that will only give us one value of x but x has two values. This is because x has a power of \[2\] so it will have two roots. We have to remember that any polynomial of n degree will exactly have n roots. We can also use the hit and trial method to solve this problem. We simply need to substitute the value in the given options in the equation and check if it satisfies the condition or not.