Question

$x+y+z=15$, $xy+yz+xz=72$, then then $3\le x\le 7$.

• A. True
• B. False

Answer 1

The correct option is A True

xy + yz + zx = 72

use: (a + b + c)^2 = a² + ² + ² + 2(ab + bc + ca), then:

x² + y² + z² = (x + y + z)² - 2(xy + yz + zx)

x² + y² + z² = 15² - 2(72) = 225 - 144 = 81

now if x = 3 , y = 6 , z = 6 or x = 6 , y = 3 , z = 6 or x = 6 , y = 6 , z = 3
check: 3 + 6 + 6 = 15 , 9 + 36 + 36 = 81

if x = 4 , y = 4 , z = 7 or x = 4 , y = 7 , z = 4 or x = 7 , y = 4 , z = 4
$\text{Thus possible values of x are {3,4,6,7}}$

So given statement is true.