Question

One number is 7 more than the other number, and its square is 77 more than the square of the smaller number. What are the numbers?

Answer 1

Hint: Assume that the numbers are x and x+7. Use the fact that the square of the larger number is 77 more than the square of the smaller number to form the equation 14x+49 = 77. Solve the equation for x and hence find the numbers.

Complete step-by-step answer:

Let the smaller of the two numbers be x.
Since the larger number is 7 more than, the smaller number, we have the other number is x+7
Now, we have
Square of the larger number $={{\left( x+7 \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above identity, we get
Square of the larger number $={{x}^{2}}+14x+49$
Also, the square of the smaller number $={{x}^{2}}$
We know that the square of the larger number is 77 more than the square of the smaller number
Hence, we have ${{x}^{2}}+14x+49={{x}^{2}}+77$
Subtracting ${{x}^{2}}$ from both sides, we get
$14x+49=77$
Subtracting 49 from both sides, we get
14x=28
Dividing both sides by 14, we get
x =2
Hence the smaller of the numbers is 2, and the larger of the two numbers is 2+7 = 9
Hence the numbers are 2 and 9

Note: Verification:
We have 9 is 7 more than 2
Also ${{9}^{2}}=81$ and ${{2}^{2}}=4$
Clearly, 81 is 77 more than 4.
Hence, the square of 9 is 77 more than the square of 4.
Hence our answer is verified to be correct.