Total no. of real roots of the equation
3x^4+4x^3+5x+1=0 are
OPtion
1) 0
2) 1
3) 2
4) 3
5) 4
6) 5
7) none of these
Solution
In 3x^4+4x^3+5x+1 no sign change so no. of positive roots = 0
now put -x in the place of x, we get
3x^4-4x^3-5x+1, this equation has sign change twice so no. of negative roots = 2
so total real roots = 0+2 = 2