Find the values of Y for real x Y= x2+2x-11 2(x-3) Share with your friends Share 0 Ajanta Trivedi answered this the given equation is : y=x2+2x-112(x-3)⇒2xy-6y=x2+2x-11⇒x2+2x.(1-y)+6y-11=0for x∈R, D≥04.(1-y)2-4*(6y-11)≥0(1-y)2-6y+11≥01+y2-2y-6y+11≥0y2-8y+12≥0y2-6y-2y+12≥0(y-6)(y-2)≥0⇒y∈(-∞,2]∪[6,∞) hope this helps you 3 View Full Answer
