Question

two candidates attempts to solve a quadratic equation of form \(x^2 px q=0\) one starts with a wrong value of p and find the roots to be 2 and 6 the other starts with a wrong value of q and finds the roots to be 2, -9 the correct roots are :

Answer 1

Hello,

Nice like problem.

(x-a)(x-b)=x²-(a+b)x+ab=x²-Sx+P

1) S=8,P=12 value of P is correct

2) S=-7, P=-18, Value of S is correct

x²-7x+12=0
Δ=7²-4*12=1
x=(7-1)/2 =3
or x=(7+1)/2=4

The correct roots are 3 and 4