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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Jan 26, 2012

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Here's my solution

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- Feb 14, 2012

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Hi Jester, thanks for participating in this problem and I really like your solution!

I'd like to share another method (which isn't my solution) on how to tackle this problem too...here goes:

We are given

\(\displaystyle 2a^2-3a-1=0\) and \(\displaystyle b^2+3b-2=0\)

We see that we can do some algebraic manipulation to the second equation to make it be another quadratic equation that is similar to that of first equation:

\(\displaystyle \frac{b^2}{b^2}+\frac{3b}{b^2}-\frac{2}{b^2}=\frac{0}{b^2}\)

\(\displaystyle 1+\frac{3}{b}-\frac{2}{b^2}=0\)

\(\displaystyle 2\left(\frac{1}{b^2}\right)-3\left(\frac{1}{b}\right)-1=0\)

Therefore, \(\displaystyle a\) and \(\displaystyle \frac{1}{b}\) are the roots for the equation \(\displaystyle 2x^2-3x-1=0\) and the intended expression could be found by finding the sum of roots and product of roots of the equation above as follows:

\(\displaystyle a+\frac{1}{b}=\frac{3}{2}\)

\(\displaystyle a\left(\frac{1}{b}\right)=-\frac{1}{2}\)

\(\displaystyle \therefore a+\frac{1}{b}+a\left(\frac{1}{b}\right)=\frac{ab+b+1}{b}=\frac{3}{2}-\frac{1}{2}=1\)

- Jan 26, 2012

- 183

Definitely cleaner than mine.

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- Feb 14, 2012

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I say both are cool and clever methods!

- Jan 25, 2013

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\(\displaystyle 2a^2-3a-1=0---(1)\) and \(\displaystyle b^2+3b-2=0---(2)\), and \(\displaystyle ab\ne 1\),

$(1)\times b+(2)\times a :$ rearrange and we get :

$(ab-1)(b+2a)=0$

$\because ab\neq 1 \therefore b=-2a$

\(\displaystyle \dfrac{ab+a+1}{b}=\dfrac{-2a^2+3a+1-2a}{-2a}=1\)

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