### Instruction: Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.### Response: This looks like a quadratic equation in disguise, since there is a square root and its reciprocal.
To make it easier to see the quadratic structure, I will 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 rid of the fraction and obtain \[y^2+10=7y.\]
This is a quadratic equation in y, so I will rearrange it in standard form as \[y^2-7y+10=0.\]
To solve for y, I will try to factor this quadratic by looking for two numbers that multiply to 10 and add to -7.
I notice that -5 and -2 are such numbers, so I can write \[y^2-7y+10=(y-5)(y-2)=0.\]
By the zero product property, this means that either y-5=0 or y-2=0, so y=5 or y=2.
But y is not the final answer, since I defined y as \sqrt{4x-3}.
To find x, I need to undo the square root by squaring both sides of the equation y=\sqrt{4x-3}.
For y=5, I get \[5^2=4x-3\] \[25=4x-3\] \[28=4x\] \[x=7.\]
For y=2, I get \[2^2=4x-3\] \[4=4x-3\] \[7=4x\] \[x=\frac{7}{4}.\]
These are the two solutions for x that satisfy the original equation.
To check, I can plug them back into the equation and see if they make it true.
For x=7, I get \[\sqrt{4(7)-3}+\frac{10}{\sqrt{4(7)-3}}=7\] \[\sqrt{25}+\frac{10}{\sqrt{25}}=7\] \[5+\frac{10}{5}=7\] \[5+2=7\] \[7=7.\]
This works, so x=7 is a valid solution.
For x=\frac{7}{4}, I get \[\sqrt{4(\frac{7}{4})-3}+\frac{10}{\sqrt{4(\frac{7}{4})-3}}=7\] \[\sqrt{7-3}+\frac{10}{\sqrt{7-3}}=7\] \[\sqrt{4}+\frac{10}{\sqrt{4}}=7\] \[2+\frac{10}{2}=7\] \[2+5=7\] \[7=7.\]
This also works, so x=\frac{7}{4} is another valid solution.
Therefore, the final answer is x=7, \frac{7}{4}.