Find sum of the roots of $$4(x-\sqrt x)^2-7x+7\sqrt x=2$$

By substituting $t=x-\sqrt x$ we have $4t^2-7t-2=0$ $$4t^2-8t+t-2=0$$ $$(4t+1)(t-2)=0$$ So we get $x-\sqrt x=2$ Hence $x=4$.

Or $x-\sqrt x=-\frac14$ then $x-\sqrt x+\frac14=0$ and $(\sqrt x-\frac12)^2=0$ and $x=\frac14$

So sum of the roots is $4+\frac14=\frac{17}{4}\quad$ Or $\quad 4+\frac14+\frac14=\frac92$ (adding $\frac14$ twice)?

  • $\begingroup$ Why would you add $\frac 14$ twice? $\endgroup$
    – abiessu
    May 11 at 19:54
  • $\begingroup$ @abiessu Because it is double root of the equation $(\sqrt x-\frac12)^2=0$. $\endgroup$
    – Soheil
    May 11 at 19:55
  • $\begingroup$ But that being a double root of that equation does not guarantee that it is a double root of the original. Otherwise you might consider adding $4$ twice for the same reason. $\endgroup$
    – abiessu
    May 11 at 19:57
  • 1
    $\begingroup$ From my understanding, the concept of the multiplicity of a root in this context is confined to polynomials. Our equation here isn't a polynomial, so I don't think the concept of the multiplicity of a root is well-defined in the typical sense. (although I could be wrong) $\endgroup$ May 11 at 20:00
  • 2
    $\begingroup$ @abiessu: In this equation $4$ is not a double root. Also, the same factoring can be done without using the substitution. $\endgroup$
    – Vasya
    May 11 at 20:01

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