Question Number 100146 by ajfour last updated on 25/Jun/20

Commented bybemath last updated on 25/Jun/20

      t = ((c^2 −2b−1)/(2(b+1)))

Commented byajfour last updated on 25/Jun/20

clearly  (√(b+t))=(c/t)  or   (b+t)t^2 = c^2   let me check     (b+((c^2 −2b−1)/(2(b+1))))[((c^2 −2b−1)/(2(b+1)))]^2             ≠ c^2   please check your answer sir,  or post your workings..

Commented byajfour last updated on 25/Jun/20

Find t in terms of b, c .

Answered by mr W last updated on 25/Jun/20

(y/1)=((b+t)/y)=(c/t)  ⇒((c/t))^2 =b+t  ⇒t^3 +bt^2 −c^2 =0  ...

Commented byajfour last updated on 25/Jun/20

thanls sir, i hope someday  some geometry shall pave the  way..!

Commented bybemath last updated on 25/Jun/20

where becomes y?

Commented by1549442205 last updated on 25/Jun/20

Yes,sir Mr.W .Please,let me be  continued.  Putting x=(1/t) we get c^2 x^3 −bx−1=0  ⇔(cx)^3 −bcx−c=o.Put cx=u we get  u^3 −bu−c=0 (1).we find the roots of   (1) in the form:u=−(m+n).Then  u+m+n=0⇒u^3 +m^3 +n^3 −3umn=0(2)  From (1) and (2) we get  { ((m^3 +n^3 =−c)),((mn=(b/3))) :}  ⇒(m^3 −n^3 )^2 =(m^3 +n^3 )^2 −4m^3 n^3 =c^2 −((4b^3 )/(27))  We need to have the condition that  c^2 −((4b^3 )/(27))≥0⇔c≥((2b)/9)(√(3b)).Then we get   { ((m^3 +n^3 =−c)),((m^3 −n^3 =(√(c^2 −((4b^3 )/(27)))))) :}      ⇒ { ((m= ^3 (√((−c+(√(c^2 −((4b^3 )/(27)))))/2)))),((n= ^3 (√((−c−(√(c^2 −((4b^3 )/(27)))))/2)))) :}  ⇒u=−(m+n)= ^3 (√((c−(√(c^2 −((4b^3 )/(27)))))/2))+ ^3 (√((c+(√(c^2 −((4b^2 )/(27)))))/2))  t=(1/x)=(c/u)=(c/(3(√((c−(√(c^2 −((4b^3 )/(27)))))/2))+ ^3 (√((c+(√(c^2 −((4b^3 )/(27)))))/2))))

Commented bymr W last updated on 25/Jun/20

thank you, very nice sir!  i have had applied cardano directly.

Commented bymr W last updated on 25/Jun/20

Commented byajfour last updated on 25/Jun/20

thanks  ....205 Sir, but cannot  we have an answer for any  real b amd c ?

Commented by1549442205 last updated on 26/Jun/20

There is always exists some root for   the values of b and c so that c^2 ≥((4b^3 )/(27)).  Example,b=3,c=4⇒t≈1.821640

Commented by1549442205 last updated on 30/Jun/20

But I want(please get his permission)   to do the his work   again in...the moment!.!.!