The product of two no. is 12. If their sum added to the sum of their squares is 32, find the no.
Answers
Answer:
3,4
Step-by-step explanation:
Let the numbers be a & b
Given,
a*b = 12
a+b + (a^2+b^2) = 32
.............................................
(a^2+b^2) = (a+b)^2 - 2*a*b [identity]
(a+b) + (a+b)^2 - 2*a*b = 32
(a+b) + (a+b)^2 - 2*12 = 32
(a+b) + (a+b)^2 = 56
(a+b)^2 + (a+b) - 56 = 0
(a+b)^2 + 8(a+b) - 7(a+b) - 56 = 0
(a+b)*[(a+b) + 8] - 7*[(a+b) + 8] = 0
[(a+b) +8]*[(a+b)-7] = 0
a+b = 7
ab = 12
b = 12/a
a + 12/a = 7
a^2 + 12 = 7a
a^2 - 7a + 12 = 0
(a-4)*(a-3) = 0
a=4 or 3
So if a = 3, b=4
Hence numbers are 3,4
Step-by-step explanation:
Product of numbers = 12
Possibilities (1,12) , (2,6), (3,4)
Sum of number + sum of squares = 32
Hence 1st and second combinations aren't possible.
So the numbers are 3,4
--------------------------------------------------------------------------------------------
Let the numbers be a & b
Given,
a*b = 12
a+b + (a^2+b^2) = 32
.............................................
(a^2+b^2) = (a+b)^2 - 2*a*b [identity]
(a+b) + (a+b)^2 - 2*a*b = 32
(a+b) + (a+b)^2 - 2*12 = 32
(a+b) + (a+b)^2 = 56
(a+b)^2 + (a+b) - 56 = 0
(a+b)^2 + 8(a+b) - 7(a+b) - 56 = 0
(a+b)*[(a+b) + 8] - 7*[(a+b) + 8] = 0
[(a+b) +8]*[(a+b)-7] = 0
a+b = 7
ab = 12
b = 12/a
a + 12/a = 7
a^2 + 12 = 7a
a^2 - 7a + 12 = 0
(a-4)*(a-3) = 0
a=4 or 3
So if a = 3, b=4
Hence numbers are 3,4
Hope it helps.