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There are constants \(\alpha\) and \(\beta\) such that \(\frac{x-\alpha}{x+\beta} = \frac{x^2-80x+1551}{x^2+57x-2970}\). What is \(\alpha+\beta\)?

 Aug 20, 2018
 #1
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There are constants  \(\alpha\) and \(\beta\) such that \(\dfrac{x-\alpha}{x+\beta} = \dfrac{x^2-80x+1551}{x^2+57x-2970}\).
What is \(\alpha+\beta\) ?

 

1.

\(\begin{array}{|rcll|} \hline && x^2-80x+1551 \\ &=& (x-40)^2 - 40^2+1551 \\ &=& (x-40)^2 - 1600+1551 \\ &=& (x-40)^2 - 49 \\ &=& (x-40)^2 - 7^2 \\ &=& [(x-40)-7][(x-40)+7] \\ &=& (x-47)(x-33) \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline && x^2+57x-2970 \\ &=& \left( x+\frac{57}{2} \right)^2 - \left(\frac{57}{2}\right)^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 28.5^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 3782.25 \\ &=& \left( x+28.5 \right)^2 - 61.5^2 \\ &=& [(x+28.5)-61.5][(x+28.5)+61.5] \\ &=& (x-33)(x+90) \\ \hline \end{array}\)

 

3.

\(\begin{array}{|rcll|} \hline && \dfrac{x^2-80x+1551}{x^2+57x-2970} \\\\ &=& \dfrac{(x-47)(x-33)}{(x-33)(x+90)} \\\\ &=& \dfrac{x-47}{x+90} \\ \hline \end{array} \)

 

4.

\(\begin{array}{|rcll|} \hline \dfrac{x-\alpha}{x+\beta} &=& \dfrac{x-47}{x+90} \\ \\ \hline x-\alpha &=& x-47 \\ -\alpha &=& -47 \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{47} \\ \\ \hline x+\beta &=& x+90 \\ \mathbf{\beta} & \mathbf{=} & \mathbf{90} \\ \\ \hline \alpha + \beta &=& 47+90 \\ &=& 137 \\ \hline \end{array}\)

 

\(\mathbf{\alpha+\beta =137}\)

 

laugh

 Aug 21, 2018
 #2
avatar+973 
0

Thanks! Your correct!

Lightning  Aug 21, 2018
 #4
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Well done! Assuming you did that by hand it was quite an achievement! However, the fact that the question only asks for \(\alpha+\beta\) suggests that the questioner probably had a short cut in mind, like the one in my answer below.

Guest Aug 21, 2018
 #3
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Since the left hand side is the ratio of two linear terms the right hand side must be too. In other words the numerator and denominator on the right hand side must contain a common factor, say \((x-\gamma)\) which cancels out. The right hand side can therefore be written:

\(\displaystyle{(x-\alpha)(x-\gamma)\over(x+\beta)(x-\gamma)} ={x^2-(\alpha+\gamma)x+\alpha\gamma\over x^2+(\beta-\gamma)x -\beta\gamma} ={x^2-80x+1551\over x^2+57x-2970}\),

so equating coefficients of x in the polynomials we must have \(\alpha+\gamma=80\) and \(\beta-\gamma=57\).

Adding these two expressions together \(\gamma\) cancels and we have immediately:

\(\alpha+\beta=80+57=137\).

 Aug 21, 2018
edited by Guest  Aug 21, 2018
 #5
avatar+973 
0

Thanks! Your correct!

Lightning  Aug 25, 2018

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