Show that the polynomial f(x)= has no zero.
Asked by araima2001 | 10th Jun, 2015, 12:59: PM
Subsitute x2=t in the polynomial to get
f(t)=t2+4t+6, a quadratic polynomial in 't'
Discriminant D of the above quadratic is 42 - 4 X 1 X 6 = 16 - 24 = -8
Since, discriminant of the quadratic is negative, the quadratic has no real roots.
Hence, for no real value of 't', the equation t2+4t+6=0 is satisfied. So, the equation has no zeroes.
Answered by satyajit samal | 10th Jun, 2015, 03:50: PM
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