The solution is very easy. At once we can see that x, y must be (1, 4) or (4, 1). Let's do it algebraically: If we have √x = a; √y = b a + b = 3 a^3/b + b^3/a = 17/2 Hence ab = 2 a+b =3 from Vieta's formula ------------------> a = x1 and b = x2 x1 + x2 = a + b x1 * x2 = ab x^2 - 3x +2 = 0 (x - 1) (x - 2) = 0 x1 = 1 x2 = 2 Hence x, y = (1, 4) or (4, 1)
*=read as square root ^=read as to the power As per question (X.*x/*y)+y.*y/*x)=17/2 {(X..*x.*x)+(y.*y.*y)}/*x.*y=17/2 (X^2+y^2)/*x.*y=17/2 If p/q=3/4 then p=3 & q=4 Similarly... X^2+y^2=17.....eqn1 *x.*y=2....eqn2 Squaring eqn2 we shall get X. Y=4.....eqn3 X+y=*{x^2+y^2+2xy} =*{17+(2×4)}=*(17+8) *25=±5.......eqn4 X-y=*{x^2+y^2-2xy} =*{17-(2×4)}=*(17-8) =*9=±3..... Eqn5 Add eqn4 & eqn5 X+y+x-y=5+3 2x=8 X=4 Put the value of x in eqn4 X+y=5 4+y=5 Y=5-4=1 Hence x=4, y=1 N:B= As per question *x+*y=3, So we can't take the negative value of (x+y)&(x-y)
Why do you waste your time to show two complex solutions? At the begining there is the domain: x, y E R+. The solution is very easy. At once we can see that x, y must be (1, 4) or (4, 1). Let's do it algebraically: If we have √x = a; √y = b a + b = 3 a^3/b + b^3/a = 17/2 Hence ab = 2 a+b =3 from Vieta's formula ------------------> a = x1 and b = x2 x1 + x2 = a + b x1 * x2 = ab x^2 - 3x +2 = 0 (x - 1) (x - 2) = 0 x1 = 1 x2 = 2 Hence x, y = (1, 4) or (4, 1) There is a saying: "A good mathematician is a lazy mathematician".