can anybody solve this question find two numbers such that their sum, their differences, and sum of their squares are in the ratio 3:1:40
Answers
Let the two numbers be x and y and the common ratio be z.
(i) Given that sum of two numbers is 3.
= > x + y = 3z ----- (1)
(ii) Given that differences of two numbers is 1
= > x - y = z ---- (2)
(iii) Given that sum of their squares is 40.
= > x^2 + y^2 = 40z ------ (3)
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On solving (1) & (2), we get
= > x + y = 3z
= > x - y = z
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2x = 4z
x = 2z.
Then substitute x = 2z in (2), we get
= > x - y = z
= > 2z - y = z
= > -y = -z
= > y = z.
hence, x = 2z and y = z.
Now,
Substitute x = 2z & y = z in (3), we get
= > x^2 + y^2 = 40z
= > (2z)^2 + (z)^2 = 40z
= > 4z^2 + z^2 = 40z
= > 5z^2 = 40z
= > z = 8
Then,
x = 2z = 2(8) = 16
y = 8.
Therefore, the two numbers are 16 and 8.
Hope this helps!