When does x+x2−1x−x2−1+x−x2−1x+x2−1=2mx+4 have 2 real solutions?

Question

Kiera Tran · Accepted Answer

a+ba−b+a−ba+b&amp;nbsp;                                                      a              +              b                                      a              −              b                                +                                    a              −              b                                      a              +              b                                                          =                                    (              a              +              b              )              (              a              +              b              )                                      (              a              −              b              )              (              a              +              b              )                                +                                    (              a              −              b              )              (              a              −              b              )                                      (              a              −              b              )              (              a              +              b              )                                                                              =                                                    a                2                            +              2              a              b              +                              b                2                            +                              a                2                            −              2              a              b              +                              b                2                                                                    a                2                            −                              b                2                                                                                            =                                    2              (                              a                2                            +                              b                2                            )                                                      a                2                            −                              b                2                                                        and this expression, in your case, when&amp;nbsp;a=x&amp;nbsp;and&amp;nbsp;b=x2−1, simplifies to something rather simple.

Jaydan Russell · Accepted Answer

After getting 2x2−mx−3=0 (I skipped the derivation because you can do it by yourself and it has also been explained by the other existing answer), we can isolate m as follows.m=2x2−3x=2x−3xwhere x≠0.- m′(x)=2+3x2&amp;gt;0 for all x∈R suggests that m(x) has no extremum. It spans from −∞ to ∞.- m(−x)=−m(x) suggests that m(x) is an odd function.- Every horizontal line y=k will cut m(x) at exactly two points.Thus for any m there are exactly two real x.

When does \(\displaystyle{\frac{{{x}+\sqrt{{{x}^{{2}}-{1}}}}}{{{x}-\sqrt{{{x}^{{2}}-{1}}}}}}+{\frac{{{x}-\sqrt{{{x}^{{2}}-{1}}}}}{{{x}+\sqrt{{{x}^{{2}}-{1}}}}}}={2}{m}{x}+{4}\) have 2 real solutions?

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