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Math Challenge 3. Chapter 5

 
 
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Math Challenge 3. Chapter 5
by Xinxin Fang - Sunday, October 11, 2020, 10:16 AM
 

I'm having trouble on 5.17. I tried rationalizing but that didn't really work. So I'm not sure if Im supposed to do some kind of substitution.\

5.20 and 5.23: I did similar things for both of these. I tried to get rid of denominators. For 5.20 I did a y substitution for x^2 but that didn't really work.

5.24: I'm not sure where to start.

 
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Re: Math Challenge 3. Chapter 5
by Dr. Kevin Wang - Monday, October 12, 2020, 2:07 AM
 

This is for MC III Algebra in the AIME Intensive Prep Class.

Question 5.17: Solve the equation:  \[ \frac{x-7}{\sqrt{x-3}+2} + \frac{x-5}{\sqrt{x-4}+1} = \sqrt{10}. \]

Rationalizing should work.  Try again, and use the substitutions $u=\sqrt{x-3}$, $v=\sqrt{x-4}$. (After rationalizing the denominators, you don't really need the substitutions to solve it, but it is cleaner to use them.)

Question 5.20:\[ \frac{1}{x^2+1} +\frac{x^2+1}{x^2} = \frac{10}{3x}. \]

Try multiplying $x$ on both sides.

Question 5.23:  \[ \frac{2x}{3} = \frac{x^2}{12}+\frac{3}{x^2}+\frac{4}{x}. \]

You can multiply $12$ on both sides and see what happens.

Question 5.24: Assume $a<30$ is an integer, and the equation (in $x$) $$\sqrt{2x-4}-\sqrt{x+a}=1$$ has exactly one integer root. Find all possible values for $a$.

There is not really a special trick for this one; you can clear the radicals first, and you get a quadratic equation in $x$, then apply the quadratic formula.  Since there is one integer root, some analysis can be done with the root.