### 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 I notice that the term under the square root appears in both terms on the left side.
I wonder what would happen if I let \(u=\sqrt{4x-3}\), then the equation would become \(u+\frac{10}{u}=7\).
That looks much simpler to work with, so I will try to solve for \(u\) first.
To get rid of the fractions, I can multiply both sides by \(u\), and get \(u^2+10=7u\).
Then I can rearrange this to get \(u^2-7u+10=0\).
This looks like a factorable quadratic, so I will try to find two numbers that multiply to \(10\) and add to \(-7\).
I can see that \(-5\) and \(-2\) work, so I can write \(u^2-7u+10=(u-5)(u-2)\).
Therefore, \(u-5=0\) or \(u-2=0\), so \(u=5\) or \(u=2\).
But I still need to find \(x\), and I remember that \(u=\sqrt{4x-3}\).
So I can plug in \(u=5\) and get \(\sqrt{4x-3}=5\).
Squaring both sides, I get \(4x-3=25\).
Adding \(3\) to both sides, I get \(4x=28\).
Dividing both sides by \(4\), I get \(x=7\).
I can do the same thing for \(u=2\) and get \(\sqrt{4x-3}=2\).
Squaring both sides, I get \(4x-3=4\).
Adding \(3\) to both sides, I get \(4x=7\).
Dividing both sides by \(4\), I get \(x=\frac{7}{4}\).
So I have found two possible values of \(x\), but I need to check if they both satisfy the original equation.
I can plug in \(x=7\) and get \(\sqrt{4(7)-3}+\frac{10}{\sqrt{4(7)-3}}=\sqrt{25}+\frac{10}{\sqrt{25}}=5+\frac{10}{5}=7\), so this works.
I can plug in \(x=\frac{7}{4}\) and get \(\sqrt{4(\frac{7}{4})-3}+\frac{10}{\sqrt{4(\frac{7}{4})-3}}=\sqrt{4}+\frac{10}{\sqrt{4}}=2+\frac{10}{2}=7\), so this also works.
Therefore, I have two solutions, \(x=7\) and \(x=\frac{7}{4}\).