# 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? Commented May 11, 2021 at 19:54
• @abiessu Because it is double root of the equation $(\sqrt x-\frac12)^2=0$.
– User
Commented May 11, 2021 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. Commented May 11, 2021 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) Commented May 11, 2021 at 20:00
• @abiessu: In this equation $4$ is not a double root. Also, the same factoring can be done without using the substitution. Commented May 11, 2021 at 20:01

We are asked to find the sum of roots of this function:

$$4(x-\sqrt{x})^2-7x+7\sqrt{x}=2 \Leftrightarrow \\ 4(x-\sqrt{x})^2-7x+7\sqrt{x}-2=0$$

$$\text{As you are saying by substituting } t=x-\sqrt{x} \text{ (1)}\text{ ,we have:}$$
$$4t^2-7t-2=0 \Leftrightarrow (4t+1)\cdot(t-2)=0$$ $$\text{And the solution we get from the previous equation are:}$$ $$\boxed{t=-\frac 1 4 \text{ or } t=2 \text{ (2)}}$$ $$\text{And now by replacing the varialbe } t \text{ we have that:}$$ $$x-\sqrt{x}=-\frac 1 4 \text{ (3) or }x-\sqrt{x}=2 \text{ (4)}$$ $$\text{Let } u=\sqrt{x} \Rightarrow u^2=x ,u \ge 0$$

$$\text{We demand } u\ge 0 \text{ because the function:} f(x)=\sqrt{x} \text{ with a domain: }$$ $$D=[0,+\infty) \text{ it has a range: } R_f=[0,+\infty)$$

$$\text{For the equation (3) we have:}$$ $$u^2-u-2=0 \Rightarrow u_{1}=2 \text{ or } u_2=-1$$
$$\text{The second root we found we have to reject it because we demand } u\ge0$$ $$\text{So we can accept only } u_1=2$$

$$\text{For the equation (4) we have:}$$ $$u^2-u+\frac 1 4=0, \Delta=0 \Rightarrow u_3=\frac {b \pm \sqrt{\Delta}} a= \frac 1 2$$

$$\text{Now by replacing the variable u we have that the roots are:}$$ $$\boxed{x_1=4 \text{ or } x_2=\frac 1 4}$$ $$\text{And now when it comes to the sum of roots we have:}$$ $$\boxed{S=x_1+x_2=4+\frac 1 4=\frac {17} {4}}$$

As you can see i didn't add twice the number $$\displaystyle\frac 1 4$$.It may be called "doulbe root" but as you can see above the quadratic formula give us one solution when $$\Delta=0$$