A Nice Algebra Problem | x + 4 = (y - 2)², y + 4 = (x - 2)² | Find the Value of x² + y²

[@azumamurakami7842]

x=0,y=0 ----> x²+y²=0 x=5,y=5 ----> x²+y²=50 x+y=3 -----> x²+y²=15

[@nasrullahhusnan2289]

x+4=(y-2)² --> x=(y-2)²-4 (1) y+4=(x-2)² --> y=(x-2)²-4 (2) From (1) note that x is a function of y. From (2) y is a function of x. Let the two functions respectively be f(y) and g(x) f(y)=(y-2)²-4 and g(x)=(x-2)²-4 (x,y) satisfying (1) and (2) is the coordinate f(y) and g(x) cross. (x-2)²-4 =0 --> x-2=±2 x={0,4} --> g(x)=0 g(.) crosses the horizontal axis at (0,0) and (4,0) Similarly, y={0,4} --> f(y)=0 f(.) crosses the vertical axis ar (0,0) and (0,4) Therefore f(y) and g(x) cross each other at (0,0). x²+y²=0

[@sunil.shegaonkar1]

Three solutions: 0, 50 & 15.

[@nasrullahhusnan2289]

Want to find x²+y² given • x+4=(y-2)² and y+4=(x-2)² • x and y are real and x≠y x-2=a and y-2=b a+b=-1 --> x+y=3 ab=-5 --> (x-2)(y-2)=-5 The solution: x²+y²=15. Why no conditions is mentioned? Otherwise it will mislead viewers to think that the solution is a circle with (0,0) as the center and radius r=sqrt(15), excluding the points that x=y. For example x=sqrt(6) --> y=±3. Using 1st given equation: LHS=x+4=4+sqrt(6) RHS=(y-2)²=1 Using 2nd given equation LHS=y+4=4±3 RHS=(x-2)² =[sqrt(6)-2]² =8-4sqrt(6) Clearly the two are not equal.

[@walterwen2975]

A Nice Algebra Problem: x + 4 = (y - 2)², y + 4 = (x - 2)², x, y ϵR, x ≠ y; x² + y² = ? (x + 4) + (y + 4) = x + y + 8 = (y - 2)² + (x - 2)² = x² + y² - 4(x + y) + 8 x + y + 8 = x² + y² - 4(x + y) + 8; x² + y² = 5(x + y) (x + 4) - (y + 4) = x - y = (y - 2)² - (x - 2)² = y² - x² + 4(x - y) x² - y² - 3(x - y) = 0, (x - y)(x + y) - 3(x - y) = (x - y)(x + y - 3) = 0 x ≠ y, x - y ≠ 0; x + y - 3 = 0, x + y = 3 x² + y² = 5(x + y) = 5(3) = 15

[@johnstanley5692]

g1=x+4*y-y^2,(=0), g2=y+4*x-x^2 (=0), g3=g1-g2 = (x-y)*(x+y-3) = 0=> x=3-y (x~=y). subs 'x' into g1 to obtain y= 3/2 +/- sqrt(21)/2 x=3-y => x= 3/2 -/+ sqrt(21)/2 => x^2+y^2 = 15