the product of two numbers is 56 when their sum is added to the sum of their squares we get 128 find the no.s
Answers
Answered by
124
xy=56
x+y+x2+y2=128
(x+y)+(x+y)^2 - 2xy=128
lets assume x+y=k
k+k^2-2(56)=128, from xy=56
k^2 +k-240=0
k=15,-16
so k=15
x+y=15
xy=56
x=8,y=7
x+y+x2+y2=128
(x+y)+(x+y)^2 - 2xy=128
lets assume x+y=k
k+k^2-2(56)=128, from xy=56
k^2 +k-240=0
k=15,-16
so k=15
x+y=15
xy=56
x=8,y=7
Answered by
41
xy = 56
x+y +x^2+y^2 +2xy = 128
( x+y) +(x+y) ×2-2xy =128
Let us assume that x+y =k
K+k × 2-2 (56)= 128, from xy =56
K×2+k - 240=0
K=15, - 16
So, k=15
x+y =15
xy = 56
x=8, y =7
Hope this answer helps you in your studies please mark it as brainlist answer
x+y +x^2+y^2 +2xy = 128
( x+y) +(x+y) ×2-2xy =128
Let us assume that x+y =k
K+k × 2-2 (56)= 128, from xy =56
K×2+k - 240=0
K=15, - 16
So, k=15
x+y =15
xy = 56
x=8, y =7
Hope this answer helps you in your studies please mark it as brainlist answer
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