# Show that the polynomial f(x)= has no zero.

### Asked by araima2001 | 10th Jun, 2015, 12:59: PM

Expert Answer:

### Subsitute x^{2}=t in the polynomial to get
f(t)=t^{2}+4t+6, a quadratic polynomial in 't'
Discriminant D of the above quadratic is 4^{2 }- 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 t^{2}+4t+6=0 is satisfied. So, the equation has no zeroes.

^{2}=t in the polynomial to get

^{2}+4t+6, a quadratic polynomial in 't'

^{2 }- 4 X 1 X 6 = 16 - 24 = -8

^{2}+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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