### Instruction: Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.### Response: To solve this equation, I want to get rid of the square roots, so I can use some algebraic techniques.
I notice that the term $\sqrt{4x-3}$ appears in both sides of the equation, so I can let $y=\sqrt{4x-3}$ and rewrite the equation as
\[ y+\frac{10}{y}=7.\]
Now, I can multiply both sides by $y$ to get
\[ y^2+10=7y.\]
This is a quadratic equation, so I can move everything to one side and try to factor it or use the quadratic formula.
I get
\[ y^2-7y+10=0.\]
This looks like it can be factored, since the factors of 10 are 1, 2, 5, and 10, and I need two of them to add up to -7.
I try
\[ y^2-7y+10=(y-5)(y-2).\]
This works, so I can use the zero product property to find the solutions for $y$.
I get
\[ y-5=0 \text{ or } y-2=0.\]
Therefore,
\[ y=5 \text{ or } y=2.\]
But remember, these are not the solutions for $x$, they are the solutions for $y=\sqrt{4x-3}$.
So, I need to plug them back into the original equation and check if they work.
If $y=5$, then $\sqrt{4x-3}=5$, which implies that $4x-3=25$, which implies that $x=7$.
This is a valid solution, since it makes the equation true when I plug it in.
If $y=2$, then $\sqrt{4x-3}=2$, which implies that $4x-3=4$, which implies that $x=\frac{7}{4}$.
This is also a valid solution, since it makes the equation true when I plug it in.