Question

Suppose x = a and x = e are two solutions to the quadratic equations x2 + px = q. Use Viete's method to find the relationship(s) between a, e.

Answer 1

Ques x²+ px =q x² + p x-q=0 x = a je reools of ove this equation. By Nietes method , ate =- There are 2 eloctions. ae =-q K. Clue) x² + px =q 3 X+ px -4 = 0 x = a; e this equation. cuee reools of third root x = k Let the These lug Niete's method, a + kte=0 k = -cate) ak+ket ae = p-ae ate + ake q ( a ) from & 2 frose 2 & 3 Cate= ( P – ae) cate) ate ad J** (ater² = (ae-p?

from 6 & ② - Cate) = late) q+aelate) = 0 H* These are the 3 relations. Que 3 x² = qx + ? (2-93-p2 het (2=ki be the red 3 root = 0 By Niele's formulae al+ ktero k= -late) aktke + ae = -q ake = p atae @ ate lakes from o & 2) from (aten qtae ate (at Qe) - Cover - ate Categ2 = (qtae) V plates +9+ae sa = 0 ** ** from & - (ate) = Golde) So we have 3 relations P + ae late) = 0 **