### Instruction: Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.### Response: This is a quadratic equation in disguise, because I can rewrite \(\sqrt{4x-3}\) as \((4x-3)^{1/2}\).
To make it easier to work with, I will let \(y=(4x-3)^{1/2}\), so the equation becomes
\[y+\frac{10}{y}=7.\]
Now I can multiply both sides by \(y\) to get rid of the fraction:
\[y^2+10=7y.\]
Subtracting \(7y\) from both sides, I get
\[y^2-7y+10=0.\]
This is a factorable quadratic, so I can use the factoring method to find the solutions for \(y\):
\[(y-5)(y-2)=0.\]
This means that \(y=5\) or \(y=2\).
But remember, \(y=(4x-3)^{1/2}\), so I need to substitute back and solve for \(x\).
If \(y=5\), then
\[(4x-3)^{1/2}=5\]
Squaring both sides, I get
\[4x-3=25.\]
Adding \(3\) and dividing by \(4\), I get
\[x=\frac{28}{4}=\frac{7}{1}.\]
If \(y=2\), then
\[(4x-3)^{1/2}=2\]
Squaring both sides, I get
\[4x-3=4.\]
Adding \(3\) and dividing by \(4\), I get
\[x=\frac{7}{4}.\]
So I have two possible values for \(x\), but I need to check if they both satisfy the original equation.
If \(x=\frac{7}{1}\), then
\[\sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=\sqrt{4\cdot\frac{7}{1}-3}+\frac{10}{\sqrt{4\cdot\frac{7}{1}-3}}=\sqrt{25}+\frac{10}{\sqrt{25}}=5+\frac{10}{5}=7.\]
So this is a valid solution.
If \(x=\frac{7}{4}\), then
\[\sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=\sqrt{4\cdot\frac{7}{4}-3}+\frac{10}{\sqrt{4\cdot\frac{7}{4}-3}}=\sqrt{4}+\frac{10}{\sqrt{4}}=2+\frac{10}{2}=7.\]
So this is also a valid solution.