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Question Number 83420 by jagoll last updated on 02/Mar/20
find range x of function   x−4(√y) = 2(√(x−y))
findrangexoffunctionx4y=2xy
Commented by mathmax by abdo last updated on 02/Mar/20
we have 0≤y≤x   (e)⇒(x−4(√y))^2 =4(x−y) ⇒  x^2 −8x(√y)+16y =4x−4y ⇒x^2 +16y−4x+4y =8x(√y) ⇒  x^2 −4x +20y =8x(√y) ⇒(x^2 −4x+20y)^2 =64x^2 y ⇒  (x^2 −4x)^2 +2(x^2 −4x)20y +400y^2  =64x^2 y ⇒  x^4 −8x^3 +16x^2 +40x^2 y−1600 xy +400y^2 −64 x^2 y =0 ⇒  400y^2  +(40x^2 −64x^2 −1660x)y +x^4 −8x^3 +16x^2  =0 ⇒  400y^2 −1684 y +x^4 −8x^3  +16x^2 =0  Δ^′ =(842)^2 −400(x^4 −8x^3  +16x^2 ) ⇒  y =((842+^− (√(842^2 −400x^4 +8.400x^3 −16.400x^2 )))/(400))...
wehave0yx(e)(x4y)2=4(xy)x28xy+16y=4x4yx2+16y4x+4y=8xyx24x+20y=8xy(x24x+20y)2=64x2y(x24x)2+2(x24x)20y+400y2=64x2yx48x3+16x2+40x2y1600xy+400y264x2y=0400y2+(40x264x21660x)y+x48x3+16x2=0400y21684y+x48x3+16x2=0Δ=(842)2400(x48x3+16x2)y=842+8422400x4+8.400x316.400x2400
Commented by mr W last updated on 02/Mar/20
to jagoll sir:  it′s wrong!  x^2 −(4t+16)x+20t^2  = 0  is not a quadratic equation for x!  because t contains x!
tojagollsir:itswrong!x2(4t+16)x+20t2=0isnotaquadraticequationforx!becausetcontainsx!
Commented by MJS last updated on 02/Mar/20
x^2 −(4t+16)x+20t^2 =0  but t=(√(x−y)) ⇒ this is not a polynome of  2^(nd)  degree in x  1−(√5)≤t≤1+(√5)  0≤(√(x−y))≤1+(√5)  0≤x−y≤6+2(√5)  and now you know nothing about x
x2(4t+16)x+20t2=0butt=xythisisnotapolynomeof2nddegreeinx15t1+50xy1+50xy6+25andnowyouknownothingaboutx
Commented by jagoll last updated on 02/Mar/20
oo yes. i understand now. thank you  mister w and mister mjs
ooyes.iunderstandnow.thankyoumisterwandmistermjs
Answered by mr W last updated on 02/Mar/20
we have  y≥0  x≥y≥0  x≥4(√y)    let (√y)=v≥0  x−4v=2(√(x−v^2 ))  x^2 −8xv+16v^2 =4(x−v^2 )  20v^2 −8xv+x^2 −4x=0  ⇒v=((2x±(√(x(20−x))))/(10))  x(20−x)≥0  ⇒0≤x≤20      ...(i)  v≥0  ⇒2x+(√(x(20−x)))≥0 ←always true!  ⇒2x−(√(x(20−x)))≥0  ⇒5x(x−4)≥0  ⇒x≥4   ...(ii)  from (i) and (ii) we get range of x:  4≤x≤20    x^2 −4(1+2v)x+20v^2 =0  x=2(1+2v)±2(√(1+4v−v^2 ))>0  1+4v−v^2 ≥0  v^2 −4v−1≤0  ⇒2−(√5)≤v≤2+(√5)  since v≥0,  ⇒0≤v≤2+(√5)  ⇒0≤y≤(2+(√5))^2 =9+4(√5)   ...(iii)  from (iii) we get range of y:  0≤y≤9+4(√5)
wehavey0xy0x4ylety=v0x4v=2xv2x28xv+16v2=4(xv2)20v28xv+x24x=0v=2x±x(20x)10x(20x)00x20(i)v02x+x(20x)0alwaystrue!2xx(20x)05x(x4)0x4(ii)from(i)and(ii)wegetrangeofx:4x20x24(1+2v)x+20v2=0x=2(1+2v)±21+4vv2>01+4vv20v24v1025v2+5sincev0,0v2+50y(2+5)2=9+45(iii)from(iii)wegetrangeofy:0y9+45

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