## Online Course Discussion Forum

### MC III Algebra Ch. 5 Help

Hi,

Can I please have some help on 5.16 and 5.20? I tried substituting y = x^2 + 1 for 5.20 but got stuck. For 5.16 I’m completely stuck.

Thanks for the help!

For 5.20, after substituting $y=x^2+1$, have you tried clearing the denominator and factoring? It should work out pretty well. Also, here is a different method: if you multiply both sides of the original equation by $x$, what will you get? Can you make a better substitution?

For 5.16, guess the root first, then prove it is the only root (if another root is bigger or smaller than the guessed one, it leads to a contradiction).

Thanks for the reply! I'm still stuck on 5.20. I tried multiplying both sides by x but then I got stuck there.

Nvm I think I got it. Can I also please have some hints for 5.17-5.19?

5.17 and 5.18 can use change of variables: you can use new variables to represent the radicals and go from there. 5.17 may require two new variables.

5.19 just needs squaring, but you want to rearrange the terms before squaring, in order for more terms after squaring to be canceled. Try different things; there are only a limited number of things to try, and the worst case scenario is that you tried all of the ways to rearrange the terms.

Can I get some hints for 5.25,26,27,28,29,30?

For 5.27, I got \( a^2+(1+3k)b^2+(1+3/(4k))c^2+d^2 \geq 2ab+3bc+2cd \) but I don't know where to go from there.

for 5.28, I got \( \frac{a_k}{a_{k+1}}=\frac{1}{a_k+1} \) but I can't find a pattern in the recursive relation.

For the other ones I have no clue.

5.27: You want the coefficients on the left hand side to be all equal, so try to set up the inequalities with $k$ as parameter in a different way to make it possible.

5.28: The following can be useful:

\[ \frac{1}{a_k + 1} = \frac{1}{a_k} - \frac{1}{a_{k+1}}. \]5.25: try rationalizing.

5.26: Cauchy-Schwarz Inequality can help.

5.29: On the second equation, add $3$ to both sides, and you can factor.

5.30: You can multiply the three equations and get rid of the radicals.

Hi,

I still need help on 27, 28, and 30.

27: I don't know how to work out the squares.

28. I got S = \( \frac{1}{a_1}-\frac{1}{a_{91}} \) But I don't know how to find a_91.

30. I don't know what to do after multiplying them. Maybe use 4R = abc/Area or something?

I have been traveling, so responded late.

Here is a more complete solution for 27:

The following inequalities are true for all real numbers $a,b,c,d$: \[\def\arraystretch{2.2} \begin{array}{rcl} 2a^2 + \dfrac{b^2}{2} \geq 2ab, \\ \dfrac{3}{2}b^2 + \dfrac{3}{2} c^2 \geq 3bc, \\ \dfrac{c^2}{2} + 2d^2 \geq 2cd. \end{array} \] Adding the inequalities, \[ 2a^2 + 2b^2 + 2c^2 + 2d^2 \geq 2ab + 3bc + 2cd, \] and equality occurs when $b=c=2a=2d$. Therefore the minimum value $M=2$.

The coefficients were found by using the parameter $k$ like suggested before.

For 28, what you did is pretty good; to estimate $a_{91}$, the following can be done:

It is clear that $\{a_k\}$ is an increasing sequence, and for all $k\geq 1$, \[ a_{k+1} - a_{k} = a_k^2 \geq a_1^2 = \frac{1}{100}, \] hence \[ a_{91} = (a_{91} - a_{90}) +(a_{90} - a_{89}) + \cdots + (a_2 - a_1) + a_1 > 90\cdot \frac{1}{100} + \frac{1}{10} = 1. \]
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