# Find the no.s if tehre square sum is 25 times of there sum and also 50 times there difference

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a*a + b*b = 25 (a+b) = 50(a - b)

25 (a+b) = 50(a - b)

a+b = (a-b)2

a + b = 2a -2b

a = 3b

b*b + 9 b*b = 100 b

10b*b =100 b

b =10

so a =30 and b=10

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Let the numbers be a,b

⇒a²+b²=25[a+b]=50[a-b]..........................[1]

we know

[a+b]²=a²+b²+2ab

substitute[1]

⇒[a²+b²]/25=a²+b²+2ab

⇒24[a²+b²]/25=2ab

⇒12a²+12b²-25ab=0

⇒[4a-3b][3a-4b]=0

⇒a=3b/4 and b=3a/4

substitute

a=3b/4 in [1]

⇒[3b/4]²+b²=25[3b/4+b]

⇒25b²/16=25[7b/4]

⇒b=28

substitute b=28 in a=3b/4

⇒a=3[28]/4

⇒a=21

therefore a=21 and b=28

⇒a²+b²=25[a+b]=50[a-b]..........................[1]

we know

[a+b]²=a²+b²+2ab

substitute[1]

⇒[a²+b²]/25=a²+b²+2ab

⇒24[a²+b²]/25=2ab

⇒12a²+12b²-25ab=0

⇒[4a-3b][3a-4b]=0

⇒a=3b/4 and b=3a/4

substitute

a=3b/4 in [1]

⇒[3b/4]²+b²=25[3b/4+b]

⇒25b²/16=25[7b/4]

⇒b=28

substitute b=28 in a=3b/4

⇒a=3[28]/4

⇒a=21

therefore a=21 and b=28