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

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)?

• Why would you add $\frac 14$ twice? May 11 at 19:54
• @abiessu Because it is double root of the equation $(\sqrt x-\frac12)^2=0$. May 11 at 19:55
• 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. May 11 at 19:57
• 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) May 11 at 20:00
• @abiessu: In this equation $4$ is not a double root. Also, the same factoring can be done without using the substitution. May 11 at 20:01