Question

one number is 7 more than another and it's square is 77more than the square of the smaller number.qhat are the numbers?​

Answer 1

Step-by-step explanation:

Let the no be x

Case 1

X+7=0

Case 2

${x }^{2} = {2}^{2} + 77$

By. Equating the equations we get

X=9

Answer 2

Given,

One number is 7 more than another and it's square is 77more than the square of the smaller number.

To find out,

What are the numbers?

Solution:

Let the smaller number be 'X'.

and the greater number is 'X+7'.

As it is given, The square of greater number is 77 more than the square of the smaller number.

${(x + 7)}^{2} = {x}^{2} + 77$

${(x + 7)}^{2} - 77 = {x}^{2}$

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

${x}^{2} + 49 + 14x - 77 = {x}^{2}$

${x}^{2} - 28 + 14x = {x}^{2}$

${x}^{2} - {x}^{2} + 14x = 28$

$14x = 28$

$x = \frac{28}{14}$

$x = 2$

Therefore the smallest number is x = 2 and the greatest number is x+7=2+7=9.

Verification:

${(x + 7)}^{2} - 77 = {x}^{2}$

Substitute the value of x and x+7 in the above equation.

${9}^{2} - 77 = 4$

$81 - 77 = 4$

$4 = 4$

L.H.S = R.H.S

Hence proved.