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Let $a$ and $b$ be the roots of the quadratic $x^2 - 5x + 3 = 2x^2 + 14x + 8.$  Find the quadratic whose roots are 1/(a + 1) and 1/(b + 1).

 Feb 11, 2024
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Simplifying the equation, we get \({x}^{2}+19x+5=0\). We know, that \(a+b=-19, ab=5\), by Vieta's formula.

Lets see if we can use this to our advantage here. A quadratic with roots \(\frac{1}{a+1}, \frac{1}{b+1}\) will have a sum of roots of \(\frac{1}{a+1} + \frac{1}{b+1} = \frac{b+1+a+1}{(a+1)(b+1)}=\frac{a+b+2}{ab+a+b+1}\). Plugging in the values we got from Vieta's, we get \(\frac{-19+2}{5-19+1}=\frac{-17}{-13}=\frac{17}{13}\)

A quadratic with roots \(\frac{1}{a+1},\frac{1}{b+1}\), will have a product of roots of \((\frac{1}{a+1})(\frac{1}{b+1})=\frac{1}{ab+a+b+1}\). Again, plugging in the values from Vieta's, we get \(\frac{1}{5-19+1}=-\frac{1}{13}\)

 

If you're looking for a monic quadratic, a possible answer would be \({x}^{2}-\frac{17}{13}x-\frac{1}{13}\). An equalivent, non monic quadratic is \(13{x}^{2}-17x-1\). Note, there are infinitely many solutions because of different possibilites of leading coefficients.

 Feb 11, 2024

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