Problem: If x is real and 4y2+4xy+x+6=0, then the complete set of values of x for which y is real, is:
Answer Choices:
A. x≦−2 and x≧3
B. x≦2 and x≧3
C. x≦−3 and x≧2
D. −3≦x≦2
E. −2≦x≦3
Solution:
We treat the equation as a quadratic equation in y for which the discriminant D=16x2−16(x+6)= 16(x2−x−6)=16(x−3)(x+2). For y to be real D≧0. This inequality is satisfied when x≦−2 or x≧3.