Math, asked by chahat393, 11 months ago

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 Anonymous
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

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