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

Lightning Aug 20, 2018

#1**+2 **

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.**

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

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

heureka Aug 21, 2018

#3**0 **

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:

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

Guest Aug 21, 2018

edited by
Guest
Aug 21, 2018