The quadratic equation x2+px+q=0x2+px+q=0   has roots -2 and 6.  Find
(a)  the value of p and of q,
(b)  the range of values of r for which the equation x2+px+q=rx2+px+q=r has no real roots.

(x+2)(x-6)=0=x^2-4x-12.

(a) p=-4 q=-12

(b) x^2+px+q=r; complete the square: x^2+px+p^2/4 +q=r+p^2/4.

(x+p/2)^2=r+p^2/4-q. When r+p^2/4-q<0 there are no real roots.

If we plug in the values of p and q from (a) we get: r+4+12<0 so r<-16.

answered Dec 15, 2016 by Top Rated User (424,820 points)