Question

separate 18 into two parts such that twice the sum of their squares is 5 times the product.

Answer 1

Answer:

Step-by-step explanation:

Let the no.s be x and y

2(x^2+y^2)=5(xy)

x+y=18 x

Answer 2

Separate 18 into two parts such that twice the sum of their squares is 5 times the product = 12 and 6

Solution:

let the numbers be x and y

$sum = x + y = 18$

$Product = xy \\\\sum\ of\ squares = x^2 + y^2$

Given,

Twice the sum of their squares is 5 times the product

$2(x^2 + y^2) = 5xy\\\\x^2 + y^2 = \frac{5}{2} \times xy\\\\Add\ 2xy\ on\ both\ sides\\\\x^2 + y^2 + 2xy = \frac{5}{2} \times xy + 2xy\\\\Then\\\\(x+y)^2 = \frac{9}{2} xy \\\\18^2 = \frac{9}{2} xy\\\\xy = 72$

$y = \frac{72}{x} \\\\Substitute\ in\ x + y = 18\\\\x + \frac{72}{x} = 18\\\\x^2 -18x + 72 = 0\\\\(x - 12)(x - 6) = 0\\\\x = 12\\\\x = 6$

Then,

$y = \frac{72}{12} = 6\\\\or\\\\y = \frac{72}{6} = 12$

Thus the two parts are 12 and 6

Learn more:

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18. Separate 18 into two parts such that twice the sum of their squares is 5 times their product.

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