### 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 the term inside the square root is a linear expression.
To make it more obvious, I will let \[y=\sqrt{4x-3}\] and rewrite the equation in terms of y.
Then I get \[y+\frac{10}{y}=7.\]
To get rid of the fraction, I will multiply both sides by y, which gives me \[y^2+10=7y.\]
Rearranging, I get \[y^2-7y+10=0.\]
This is a quadratic equation that I can factor into \[(y-5)(y-2)=0.\]
So the possible values of y are 5 and 2, but I have to remember that y is actually the square root of something, so I have to check if these values make sense.
If y is 5, then \[y=\sqrt{4x-3}\] implies that \[5=\sqrt{4x-3}\] or \[25=4x-3\] or \[x=7.\]
This is a valid solution, because plugging in x=7 into the original equation gives me \[ \sqrt{4(7)-3}+\frac{10}{\sqrt{4(7)-3}}=7,\] which is true.
If y is 2, then \[y=\sqrt{4x-3}\] implies that \[2=\sqrt{4x-3}\] or \[4=4x-3\] or \[x=\frac{7}{4}.\]
This is also a valid solution, because plugging in x=7/4 into the original equation gives me \[ \sqrt{4(\frac{7}{4})-3}+\frac{10}{\sqrt{4(\frac{7}{4})-3}}=7,\] which is also true.
So the two solutions are x=7 and x=7/4, and I can enter them as 7,7/4.