Question

One number is 7 more than the other and its square is 77 more than the square of the smaller number

Answer 1

Solution:

Let one number = x

second number = x+7

According to the problem given,

(x+7)²-x² = 77

=> x²+2*x*7+7²-x²=77

=> x²+14x+49-x²=77

=> 14x+49 = 77

=> 14x = 77-49

=> 14x = 28

=> x = 28/14

=> x = 2

Therefore,

First number (x) = 2

Second number (x+7) = 2+7=9

••••

Answer 2

Answer:

Given that:-

• A number is 7 less than the other number.

• Its square is 77 less than the square of the greater number.

To find :-

• The Smaller Number

$\huge{\sf{\boxed{\boxed{Solution:}}}}$

= let the greater number be x

= Let the smaller number be x + 7

$\bold{A.T.Q}$

⇒ (x + 7)²  - 77 = x²

⇒ x² + 49 + 14x - 77 = x²

⇒ x² - 28 + 14x = x²

⇒ x²- x² + 14x = 28

⇒ 14x = 28

⇒ x = 28/14

⇒ x = 2

Greater number = x = 2

$\bold{Smaller\: number\: = \:x + 7 = 2 + 7 \:= 9}$