the difference between two positive whole number is 3 and the sum of their squares is 117 by calculating let us write the two number
Answers
x = 9 ; y = 6
Step-by-step explanation:
let the nos. be x and y , where x>y
x-y = 3 --------(eq. i)
x²+y²= 117 ------(eq ii)
(x-y)² = x²+y²+ 2xy
3² = 117-2xy
2xy = 117 - 9
2xy = 108
xy = 108 / 2
--》xy = 54
x = 54/y ---(iii)
substitute value of x from eq (iii) in eq (i)
54/y - y = 3
( 54 - y² )/ y = 3
54 - y² = 3y
y² + 3y - 54 = 0
y² + 9y -6y -54 =0
y(y+9) -6(y+9) =0
(y+9) (y-6)= 0
either y = -9 or y = 6 ( since x and y are whole nos. therefore negative value is not possible)
it implies that y = 6
substitute the value of y in eq (i)
x-y = 3
x-6 = 3
x = 3+6
x = 9
ans : x = 9 ; y = 6
Step-by-step explanation:
Given :-
The difference between two positive whole number is 3 and the sum of their squares is 117.
To find :-
Find the two numbers ?
Solution :-
Let the two whole numbers be X and Y
X > Y
Given that
The difference between two positive whole number = 3
=> X - Y = 3 --------(1)
=> X = 3+Y ---------(2)
and
Thee sum of their squares = 117
=> X² + Y² = 117 -------(3)
On squaring equation (1) both sides then
=> (X-Y)² = 3²
=> X²-(2X)(Y) + Y² = 9
Since (a-b)² = a²-2ab+b²
=> X²+Y²-2XY = 9 ------(4)
From (2) &(3)
=> 117- 2Y(3+Y) = 9
=> 117-6Y-2Y² = 9
=> 117-6Y-2Y²-9 = 0
=> -2Y²-6Y+108 = 0
=> -2(Y²+3Y-54) = 0
=> Y²+3Y-54 = 0/-2
=> Y²+3Y-54 = 0
=> Y²+9Y-6Y -54 = 0
=> Y(Y+9)-6(Y+9) = 0
=> (Y+9)(Y-6) = 0
=> Y+9 = 0 or Y-6 = 0
=> Y = -9 or Y = 6
Y can not be negative.
Since the given are positive whole numbers
Y = 6
On Substituting the value of Y in (2)
=> X = 6+3
=> X = 9
Therefore, X = 9 and Y = 6
Answer:-
The required positive whole numbers are 9 and 6
Check:-
The two whole numbers = 9 and 6
Their difference = 9-6 = 3
Sum of their squares
=> 9²+6²
=> 81+36
=> 117
Verified the given relations in the given problem.
Used formulae:-
- (a-b)² = a²-2ab+b²