two numbers differ by 4 and their product is 192 find the numbers
Answers
Answered by
54
Hey!!!!
Good Afternoon
__________
let the numbers be x and y
ATQ,
=> x - y = 4
=> x = 4 + y -------(1)
Also given
=> xy = 192
Substituting (1) in above equation
=> y(4 + y) = 192
=> y² + 4y = 192
By completing the square method
=> y² + 4y + (2)² = 192 + 4
=> (y + 2)² = 196
=> y + 2 = +- 14
Case 1 , y + 2 = 14
=> y = 12
Case 2, y + 2 = -14
=> y = -16
From Case 1, y = 12 and x = y + 4
=> x = 16
Thus the numbers numbers are 12 and 16.
_____________
Hope this helps ✌️
Good Afternoon
__________
let the numbers be x and y
ATQ,
=> x - y = 4
=> x = 4 + y -------(1)
Also given
=> xy = 192
Substituting (1) in above equation
=> y(4 + y) = 192
=> y² + 4y = 192
By completing the square method
=> y² + 4y + (2)² = 192 + 4
=> (y + 2)² = 196
=> y + 2 = +- 14
Case 1 , y + 2 = 14
=> y = 12
Case 2, y + 2 = -14
=> y = -16
From Case 1, y = 12 and x = y + 4
=> x = 16
Thus the numbers numbers are 12 and 16.
_____________
Hope this helps ✌️
Answered by
20
Given,
a−b=4......(1)
ab=192
Formula,
(a+b)^2+(a−b)^2=4ab
(a+b) ^2+4^2=4(192)
(a+b) ^2 =784
∴a+b=28....(2)
solving (1) and (2), we get
a=16,b=12.
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