### 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, because the square root and its reciprocal are added together.
To make it more obvious, 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 and get \[ y^2+10=7y. \]
Subtracting 7y from both sides gives me \[ y^2-7y+10=0. \]
This looks like a factorable quadratic, so I can try to find two numbers that add up to -7 and multiply to 10.
The numbers are -5 and -2, so I can write \[ y^2-7y+10=(y-5)(y-2)=0. \]
To solve for y, I can use the zero product property and set each factor equal to zero.
This gives me \[ y-5=0 \quad \text{or} \quad y-2=0. \]
Solving for y, I get \[ y=5 \quad \text{or} \quad y=2. \]
But these are not the solutions for x, I still need to substitute back the original expression for y.
So, I have \[ \sqrt{4x-3}=5 \quad \text{or} \quad \sqrt{4x-3}=2. \]
To solve for x, I can square both sides of each equation and simplify.
For the first equation, I get \[ 4x-3=25 \]
Adding 3 to both sides and dividing by 4, I get \[ x=\frac{28}{4}=\frac{7}{1}. \]
For the second equation, I get \[ 4x-3=4 \]
Adding 3 to both sides and dividing by 4, I get \[ x=\frac{7}{4}. \]
These are the two solutions for x, but I need to check if they are valid, because I squared both sides and I might have introduced extraneous solutions.
To check, I can plug them back into the original equation and see if they make it true.
For x=7/1, I get \[ \sqrt{4\cdot\frac{7}{1}-3}+\frac{10}{\sqrt{4\cdot\frac{7}{1}-3}}=\sqrt{25}+\frac{10}{\sqrt{25}}=5+2=7, \] which is true.
For x=7/4, I get \[ \sqrt{4\cdot\frac{7}{4}-3}+\frac{10}{\sqrt{4\cdot\frac{7}{4}-3}}=\sqrt{4}+\frac{10}{\sqrt{4}}=2+5=7, \] which is also true.
So, both solutions are valid, and I can write them as my final answer.