+400 Let a and b be the roots of the quadratic x2−5x+3=2x2+14x+8. Find the quadratic whose roots are 1/(a + 1) and 1/(b + 1).
+410 Simplifying the equation, we get x2+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 1a+1,1b+1 will have a sum of roots of 1a+1+1b+1=b+1+a+1(a+1)(b+1)=a+b+2ab+a+b+1. Plugging in the values we got from Vieta's, we get −19+25−19+1=−17−13=1713.
A quadratic with roots 1a+1,1b+1, will have a product of roots of (1a+1)(1b+1)=1ab+a+b+1. Again, plugging in the values from Vieta's, we get 15−19+1=−113.
If you're looking for a monic quadratic, a possible answer would be x2−1713x−113. An equalivent, non monic quadratic is 13x2−17x−1. Note, there are infinitely many solutions because of different possibilites of leading coefficients.
