If the roots of the equation (a−1)x2+x+12=(a+1)x4+x2+1are real and distinct, then the value of a∈

x4+x2+1=x2+12−x2=x2+x+1x2−x+1 x2+x+1=x+122+34≠0∀xTherefore, we can cancel this factor and we get(a−1)x2−x+1=(a+1)x2−x+1   or   x2−ax+1=0It has real and distinct roots if D=a2−4>0.