Math Olympiad | Can You Solve This Algebriac Equations | Find the Value of x and y 

Good. But, confusing when the same letter is reused.

Why do you reuse y ? Using an another letter would be better for understanding.

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)

I solved it in my mind only in 1 minute, without using pen and paper.

*=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)

If y =b2, how y=ab?

Not agree...it's so cnfusing ab=√xy होना चाहिए Why are you assuming ab=y

wrong method. you can not put y for two different values. pi. justify your stand. ndas

Math Olympiad: (x√x)/√y + y√y)/√x = 17/2, √x + √y = 3, x, y ϵR⁺ (√x + √y)² = 3², x + y + 2√(xy) = 9; √(xy) = [9 - (x + y)]/2 (x√x)/√y + y√y)/√x = [(x√x)√x + (y√y)√y)/√(xy) = (x² + y²)/√(xy) = 17/2 [2(x² + y²)]/[9 - (x + y)] = 17/2, 4(x² + y²) = 17[9 - (x + y)] = 153 - 17(x + y) 4(x² + y²) = 4(x + y)² - 8xy = 4(x + y)² - 8[9 - (x + y)]²/4 = 4(x + y)² - 2[9 - (x + y)]² = 4(x + y)² - 2[81 - 18(x + y) + (x + y)²] = 2(x + y)² + 36(x + y) - 162 = 153 - 17(x + y) 2(x + y)² + 53(x + y) - 315 = 0, [(x + y) - 5][2(x + y) + 63] = 0, x, y ϵR⁺ 2(x + y) + 63 > 0; (x + y) - 5 = 0, x + y = 5 √(xy) = [9 - (x + y)]/2 = (9 - 5)/2 = 2, xy = 4; y = 5 - x, x(5 - x) = 4 x² - 5x + 4 = 0, (x - 1)(x - 4) = 0; x - 1 = 0 or x - 4 = 0 x = 1, y = 5 - x = 5 - 1 = 4 or x = 4, y = 5 - 4 = 1 Answer check: (x√x)/√y + y√y)/√x = (x² + y²)/√(xy) = 17/2, √x + √y = 3 x = 1, y = 4 or x = 4, y = 1 (x² + y²)/√(xy) = (1 + 4²)/√4 = (4² + 1)/√4 = 17/2; Confirmed √x + √y = √1 + √4 = √4 + √1 = 3; Confirmed Final answer: x = 1, y = 4 or x = 4, y = 1

let a=sqrt(x), b=sqrt(y) then g1=a+b-3 (=0). a^3/b+b^3/a-17/2 (= 0 )=> g2=2*a^4-17*a*b+2*b^4 = 0. eliminate 'a' using g2/g1=0. i.e. remainder (g2/g1) = 4*b^4-24*b^3+125*b^2-267*b+162 = (4*b^2 - 12*b + 81)*(b^2 - 3*b + 2) = 0 Hence b={1, 2, 3/2 + 2^(1/2)*3i, 3/2 - 2^(1/2)*3i} subs 'b' into g1 => a={2, 1, 3/2 - 2^(1/2)*3i, 3/2 + 2^(1/2)*3i} =>x=a^2={ 4, 1, - 63/4 - 2^(1/2)*9i, - 63/4 + 2^(1/2)*9i }, y=b^2={ 1, 4, - 63/4 + 2^(1/2)*9i, - 63/4 - 2^(1/2)*9i }

Why Let ab=y. ab=y=b^2= root over xy