Question

Sum of all natural numbers a such that \[{a^2} - 16a + 67\] is a perfect square is
A.10
B.12
C.16
D.22

Answer 1

Hint: Here they have already given the hint in the question. We need to find the value such that the given equation is a perfect square. Now if we observe that the first two terms can be a perfect square if completed with the third term . So we will split the terms to make it a perfect square. Like 67 can be written as 64 plus 3. This will help to make the perfect square. And then we will check for the values of a.

Complete step by step solution:
Given equation is,
 \[{a^2} - 16a + 67\]
On splitting the terms as \[16a = 2 \times 8a\] and \[67 = 64 + 3\]
 \[{a^2} - 2 \times 8a + 64 + 3\]
Now we can see a perfect square of the form \[{a^2} - 2ab + {b^2}\] and that can be written as, \[{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}\]
So let’s complete the step,
 \[{\left( {a - 8} \right)^2} + 3\]
Now to make tis a perfect square we know the nearest perfect square as 4.
So the bracket value should be 1 or -1. So the value of a can be \[a = 9\;or\;a = 7\]
Now we are at the last step of the solution.
The values that make it a perfect square are 9 and 7.
So the sum will be \[9 + 7 = 16\]
So option C is the correct one.
So, the correct answer is “Option C”.

Note: Students note that we have to make the equation a perfect square and not the first two terms only. So we used splitting of the terms. Because that helps us to make a perfect square there only. So don’t be confused.
 \[{a^2} - 16a + 67\] is not a perfect square but there are values of a that gives the output as a perfect square.
For a=7 \[{a^2} - 16a + 67 = 49 - 112 + 67 = 4\]
For a=9 \[{a^2} - 16a + 67 = 81 - 144 + 67 = 4\]