Reduce the expression y=x2−x+1x2+x+1 to the form ax2+bx+c and give condition for x to be real.
(y+1)2−4(y−1)2≥0
y=x2−x+1x2+x+1 ⇒ x2(y−1)+x(y+1)+(y−1) = 0
This is in the form of ax2+bx+c=0
Condition for x to be real is D ≥ 0
⇒b2−4ac≥0
⇒(y+1)2−4(y−1)2≥0