Is value of α defined?Consider three quadratic functions: P 1 ( x ) = a x 2 − b x − c P 2 ( x ) = b x 2 − c x − a P 3 ( x ) = c x 2 − a x − bWhere a , b , c ∈ R ∖ { 0 } If there exists real number α such that P 1 ( α ) = P 2 ( α ) = P 3 ( α )Prove that a = b = cMy Try:Using P 1 ( α ) = P 2 ( α ) = P 3 ( α )we get (1) ( a − b ) α 2 − ( b − c ) α − ( c − a ) = 0 (2) ( b − c ) α 2 − ( c − a ) α − ( a − b ) = 0 (3) ( c − a ) α 2 − ( a − b ) α − ( b − c ) = 0 Letting p = a − b, q = b − c and r = c − a we get the above three equations as: p α 2 − q α − r = 0 q α 2 − r α − p = 0 r α 2 − p α − q = 0Using p + q + r = 0 and Eliminating α from last two equations above we get: (4) α = p 2 + p r + r 2 p 2 + p r + r 2 = 1 Substituting α = 1 we get: p − q − r = 0 ⟹ p = 0Similarly q = r = 0But p = r = 0 is contradicting the result in (4)So is α = 1 or α is not defined?

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Is value of   α defined?Consider three quadratic functions:      P    1    (  x  )  =  a      x    2    −  b  x  −  c      P    2    (  x  )  =  b      x    2    −  c  x  −  a      P    3    (  x  )  =  c      x    2    −  a  x  −  bWhere   a  ,  b  ,  c  ∈      R    ∖      {    0    }  If there exists real number   α such that      P    1    (  α  )  =      P    2    (  α  )  =      P    3    (  α  )Prove that   a  =  b  =  cMy Try:Using      P    1    (  α  )  =      P    2    (  α  )  =      P    3    (  α  )we get                    (1)                    (        a        −        b        )                  α          2                −        (        b        −        c        )        α        −        (        c        −        a        )        =        0                                (2)                    (        b        −        c        )                  α          2                −        (        c        −        a        )        α        −        (        a        −        b        )        =        0                                (3)                    (        c        −        a        )                  α          2                −        (        a        −        b        )        α        −        (        b        −        c        )        =        0            Letting   p  =  a  −  b,   q  =  b  −  c and   r  =  c  −  a we get the above three equations as:  p      α    2    −  q  α  −  r  =  0  q      α    2    −  r  α  −  p  =  0  r      α    2    −  p  α  −  q  =  0Using   p  +  q  +  r  =  0 and Eliminating   α from last two equations above we get:                    (4)                    α        =                                            p              2                        +            p            r            +                          r              2                                                          p              2                        +            p            r            +                          r              2                                      =        1            Substituting   α  =  1 we get:  p  −  q  −  r  =  0    ⟹    p  =  0Similarly  q  =  r  =  0But   p  =  r  =  0 is contradicting the result in (4)So is   α  =  1 or   α is not defined?

Arely Davila · Accepted Answer

Step 1The result in (4) is only valid if       p    2    +  p  r  +      r    2    ≠  0, so that you can divide by it. So there is no issue with reaching a contradiction from (4); that just is a proof by contradiction that       p    2    +  p  r  +      r    2    =  0. (Actually, there is no problem reaching a contradiction even without this comment, since that would just prove the hypotheses are impossible and so the conclusion   a  =  b  =  c is vacuously true.)Step 2To complete your solution, then, you just have to prove that   a  =  b  =  c if       p    2    +  p  r  +      r    2    =  0. This is easy:       p    2    +  p  r  +      r    2    =  0 for real p,r implies   p  =  r  =  0 (if   p  ≠  0 you can divide by       p    2   and find that r/p is nonreal, and similarly if   r  ≠  0). This then gives   a  =  b  =  c.

Jaslyn Sloan · Accepted Answer

Step 1Your main error occurred while eliminating       α    2   from equations   q      α    2    −  r  α  −  p  =  0 and   r      α    2    −  p  α  −  q  =  0.Reason:Once you eliminate       α    2   we get  (      r    2    −  p  q  )  α  =      p    2    −  q  rHere actually you have taken the term       r    2    −  p  q to the denominator without knowing its non zero nature.Now here is the solution:In case       r    2    −  p  q  =  0 we get       p    2    +  p  r  +      r    2    =  0, But since p,r are reals, this gives   p  =  0  =  r and so   q  =  0 too.So your equation (4) has no meaning, that means for any   α its always true that  (      r    2    −  p  q  )  α  =      p    2    −  q  rStep 2Now in case       r    2    −  p  q  ≠  0, then as you pointed out   α  =  1, which gives again   p  =  q  =  r  =  0 which is contradicting       r    2    −  p  q  ≠  0, hence       r    2    −  p  q should be zero.So finally       r    2    −  p  q  =  0 which gives   p  =  q  =  r  =  0 which implies   a  =  b  =  c for any   α.So essentially value of   α is irrelevant here.

Is value of alpha defined? Consider three quadratic functions: P_1(x)=ax^2-bx-c, P_2(x)=bx^2-cx-a, P_3(x)=cx^2-ax-b.

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