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x-x-4a-x-x-4a-a-0-find-x-in-terms-of-a-

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Question Number 183989 by HeferH last updated on 01/Jan/23
(((√x) + (√(x − 4a)))/( (√x) − (√(x − 4a)))) = a ≠ 0 find “x” in terms of “a”.
x+x−4ax−x−4a=a≠0find“x″intermsof“a″.
Commented by Frix last updated on 01/Jan/23
I get x=(a+1)^2
Igetx=(a+1)2
Answered by Rasheed.Sindhi last updated on 01/Jan/23
(((√x) + (√(x − 4a)))/( (√x) − (√(x − 4a)))) = a ≠ 0 find “x” in terms of “a”. (((√x) +(√(x−4a)) )/( (√x) −(√(x−4a)) ))+1=a+1 ((2(√x))/( (√x) −(√(x−4a)) ))=a+1 2(√x) =a(√x) −a(√(x−4a)) +(√x) −(√(x−4a)) a(√x) −(√x) =a(√(x−4a)) +(√(x−4a)) (√x) (a−1)=(√(x−4a)) (a+1) (((√(x−4a)) )/( (√x)))=((a−1)/(a+1)) ((x−4a)/x)=(((a−1)/(a+1)))^2 1−((4a)/x)=((a^2 −2a+1)/(a^2 +2a+1)) −((4a)/x)=((a^2 −2a+1)/(a^2 +2a+1))−1=((−4a)/((a+1)^2 )) (1/x)=(1/((a+1)^2 )) [∵ a≠0] x=(a+1)^2
x+x−4ax−x−4a=a≠0find“x″intermsof“a″.x+x−4ax−x−4a+1=a+12xx−x−4a=a+12x=ax−ax−4a+x−x−4aax−x=ax−4a+x−4ax(a−1)=x−4a(a+1)x−4ax=a−1a+1x−4ax=(a−1a+1)21−4ax=a2−2a+1a2+2a+1−4ax=a2−2a+1a2+2a+1−1=−4a(a+1)21x=1(a+1)2[∵a≠0]x=(a+1)2
Answered by a.lgnaoui last updated on 01/Jan/23
((((√x) +(√(x−4a)) )^2 )/(4a))=a 2x−4a+2(√(x(x−4a))) =4a^2 (√(x(x−4a))) =2a^2 +2a−x x(x−4a)=4a^4 +4a^2 +x^2 +8a^3 −4a^2 x−4ax x^2 −4ax=x^2 −4ax(a+1)+4a^2 (a^2 +2a+1) 0=4a^2 (a+1)^2 −4a^2 x x=(a+1)^2
(x+x−4a)24a=a2x−4a+2x(x−4a)=4a2x(x−4a)=2a2+2a−xx(x−4a)=4a4+4a2+x2+8a3−4a2x−4axx2−4ax=x2−4ax(a+1)+4a2(a2+2a+1)0=4a2(a+1)2−4a2xx=(a+1)2
Answered by manolex last updated on 01/Jan/23
m=(√x) n=(√(x−4a)) ((m+n)/(m−n))=a m^2 =x n^2 =x−4a m^2 −n^2 =4a (m+n)(m−n)=4a ((m+n)/(m−n))=a m≠n (m+n)^2 =4a^2 m+n=2a ∨ m+n=−2a m−n=2 m−n=−2 2m=2a+2 2m=−2(a+1) m=a+1 m=−(a+1) (√x)=a+1 (√x)=−(a+1) x=(a+1)^2 x=(a+1)^2 solucion x=(a+1)^2 comprobacion ((a+1+(√((a+1)^2 −4a)))/(a+1−(√((a+1)^2 −4a))))=((a+1+(a−1))/(a+1−(a−1)))=((2a)/2)=a correcto
m=xn=x−4am+nm−n=am2=xn2=x−4am2−n2=4a(m+n)(m−n)=4am+nm−n=am≠n(m+n)2=4a2m+n=2a∨m+n=−2am−n=2m−n=−22m=2a+22m=−2(a+1)m=a+1m=−(a+1)x=a+1x=−(a+1)x=(a+1)2x=(a+1)2solucionx=(a+1)2comprobaciona+1+(a+1)2−4aa+1−(a+1)2−4a=a+1+(a−1)a+1−(a−1)=2a2=acorrecto
Answered by Rasheed.Sindhi last updated on 01/Jan/23
(((√x) + (√(x − 4a)))/( (√x) − (√(x − 4a)))) = a ≠ 0 find “x” in terms of “a”. (√x) +(√(x−4a)) =a(√x) −a(√(x−4a)) a(√x) −(√x) =(√(x−4a)) +a(√(x−4a)) (√x) (a−1)=(√(x−4a)) (a+1) (((√(x−4a)) )/( (√x)))=((a−1)/(a+1)) ((x−4a)/x)=((a^2 −2a+1)/(a^2 +2a+1)) ((x−4a)/x)−1=((a^2 −2a+1)/(a^2 +2a+1))−1 ((−4a)/x)=((−4a)/((a+1)^2 )) (1/x)=(1/((a+1)^2 )) [∵a≠0] x=(a+1)^2
x+x−4ax−x−4a=a≠0find“x″intermsof“a″.x+x−4a=ax−ax−4aax−x=x−4a+ax−4ax(a−1)=x−4a(a+1)x−4ax=a−1a+1x−4ax=a2−2a+1a2+2a+1x−4ax−1=a2−2a+1a2+2a+1−1−4ax=−4a(a+1)21x=1(a+1)2[∵a≠0]x=(a+1)2

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