Divide 25 into two parts such that twice the square
of the larger part exceeds thrice the square of the smaller part by 29.
Answers
Answered by
2
Answer:
X=36
Y=-11
Step-by-step explanation:
Let 25=x+y
X=25-y
Let x be larger than y
2(x)^2-3(y)^2=29
2(25-y)^2-3(y)^2=29
2(625+y^2+50y)-3(y)^2=29
1250+2(y)^2+100y-3(y)^2=29
1250-y^2+100y=29
-y^2+100y+1250=29
By solving
Y=-11
X=36
Similar questions