Common ratio

Paul, Mike, and Kate solved the following task:

Determine the common ratio of an infinite convergent geometric sequence in which the first term is equal to $6$, and the sum of all its terms is one-eighth of the sum of squares of all these terms.

All three knew the formula for the sum of an infinite convergent geometric series with the first term $a_1$ and common ratio $q$: $$ S=\frac{a_1}{1-q} $$

Paul reasoned as follows: For the sum of squares of all terms of the given sequence, it must hold: $$ S_{\square}=\frac{36}{1-q^2 } $$ According to the task assignment, it should apply $S=\frac18 S_{\square}$. Thus, he got the equation: $$ \frac{6}{1-q}=\frac18 \cdot \frac{36}{1-q^2 } $$ Next, Paul cleared fractions from the above equation, transformed the resulting equation into the standard form of a quadratic equation, and solved it: $$ \begin{gather} 48-48q^2=36-36q \cr 4q^2-3q-1=0 \cr q_{1,2}=\frac{3 \pm \sqrt{(-3)^2-4 \cdot 4 \cdot (-1) }}{2\cdot 4} \cr q_1=1,~q_2=-\frac14 \end{gather} $$

Mike thought as follows: If $S=\frac{6}{1-q}$, then for the sum of squares of all terms of the given sequence, it should hold: $$ S_{\square}=\left(\frac{6}{1-q}\right)^2 $$ Further, he continued. If $S=\frac18 S_{\square}$ is to hold, then: $$ \frac{6}{1-q}=\frac18 \cdot \left(\frac{6}{1-q}\right)^2 $$ From the above equation, he eliminated fractions multiplying both sides by $8(1-q)^2$ and obtained: $$ \begin{gather} 48(1-q)=36 \cr q=\frac14 \end{gather} $$

Kate was convinced that the two sequences differ in the common ratio, and so the sum of squares of all the terms of the given sequence should be: $$ S_{\square}=\frac{6}{1-q^2 } $$ Then she continued in the following way: $$ \frac{6}{1-q}=\frac18 \cdot \frac{6}{1-q^2 } $$ She multiplied both sides of the equation by $8(1-q^2 )$ and got: $$ \begin{gather} 48(1+q)=6 \cr q=-\frac78 \end{gather} $$ Did any of them get the correct result?