One number is 7 more than another and it's square is 77 more than the square of the smaller number,what are the numbers?
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Udayvamsi:
Nice
but wont we get 2 cases, either ways the answer will be 2 and 9
yes i have written it in last see first
okay , just clarifying
its ok
Answered by
4
Let "x" and "y" be the unknown real numbers. (case 1,where y<x and case 2 where x<y)
As it is given in the question, we get '2' equations.
CASE 1
1) x + 7 = y (one number is 7 more than another)
2) x^2 = 77 + y^2 (square of one number is 77 more than square of smaller number)
solving 2) we get
x^2 - y^2 = 77
3) (x + y)(x - y) = 77
but from 1) we can get
x - y + 7 = 0
4) x - y = -7
substituting 4) in 3) we get
(x + y)(-7) = 77
5) x + y = -11
From 5) and 4) we can say
x = -9 and y= -2
CASE 2
y^2 = 77 + x^2 (first equation remains the same as per the question)
Therefore
(y-x)(y+x)=77 ( but as we know x-y= -7 so, y-x=7)
Hence we get,
y+x=11 and we already have y-x=7
so y=9 and x=2
so from both cases we can conclude
y= 9 or -2
and
x= 2 or -9
as the square of the numbers will be positive
x=9 y=2
As it is given in the question, we get '2' equations.
CASE 1
1) x + 7 = y (one number is 7 more than another)
2) x^2 = 77 + y^2 (square of one number is 77 more than square of smaller number)
solving 2) we get
x^2 - y^2 = 77
3) (x + y)(x - y) = 77
but from 1) we can get
x - y + 7 = 0
4) x - y = -7
substituting 4) in 3) we get
(x + y)(-7) = 77
5) x + y = -11
From 5) and 4) we can say
x = -9 and y= -2
CASE 2
y^2 = 77 + x^2 (first equation remains the same as per the question)
Therefore
(y-x)(y+x)=77 ( but as we know x-y= -7 so, y-x=7)
Hence we get,
y+x=11 and we already have y-x=7
so y=9 and x=2
so from both cases we can conclude
y= 9 or -2
and
x= 2 or -9
as the square of the numbers will be positive
x=9 y=2
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