+0

0
18
1
+400

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

#1
+394
+2

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