The sum of squares of two numbers is 208. The square of larger number is 18 times the smaller number
Answers
Answer:
x = 12 , y = 8
Solution:
Let the required numbers be x and y such that x > y .
According to the question ,
Sum of zeros of the required numbers is 208 .
Thus,
x² + y² = 208 -------(1)
Also,
The square of the larger number is 18 times of the small number .
Thus,
x² = 18y -------(2)
Now,
Putting x² = 18y in eq-(1) , we get ;
=> x² + y² = 208
=> 18y + y² = 208
=> y² + 18y - 208 = 0
=> y² + 26y - 8y - 208 = 0
=> y(y + 26) - 8(y + 26) = 0
=> (y + 26)(y - 8) = 0
Case(1) : y + 26 = 0
=> y + 26 = 0
=> y = -26 (rejected)
When we observe eq-(2) , ie ; x² = 18y
=> y = x²/18
We know that x²/18 is always positive thus y also must be positive . But we got y = -26 which is negative .
Thus,
y = -26 will be rejected .
Case(2) : y - 8 = 0
=> y - 8 = 0
=> y = 8 (appropriate value)
Now,
Putting y = 8 in eq-(2) , we get ;
=> x² = 18y
=> x² = 18×8
=> x² = 144
=> x = √144
=> x = ±12 { x = -12 is rejected }
=> x = 12 { x = 12 is appropriate , as x > y }