We all find no. of real solutions for a 2 degree equation....
but wat if the ask u like a polynomial of degree 8 , they ask u to find the no. of real roots...+ive or -ive.... or complex roots....
dont worry...herez a sol. (some of u might know ) but rest.....look here
for eg. equation is
x4 - 3x + 1 = 0....
we can write it as
(+) x4 - (3x) + (1) = 0...
see the sign change.....
+ to - , then - to + , 2 changes na
so no. of +ive solutions = 2
now to -ive solutions
we gotf(x) = x4 + 3x + 1 = 0....
take f(-x) = x4 + 3x +1 = 0.....
no sign changes......so no -ive root
for complex roots , see the highest degree.....i.e 4 here
so complex roots = highest degree - (no. of + ive + no. of -ive roots)
= 4 - (2+0) = 2 complex roots
hope u got it....ok so now post the no. of +ive , -ive n complex roots =
x8 - 4x7 + 3x6 + x5 - x4 + x3 - 16x + 2 = 0
1 thing more....
if equation of type ax2 + b |x| + c = 0 , no. of real solutions is = 0
and when roots of an equation are equal in magnitude but of opp. sign then sum of roots is = to zero......
hope it helps....
cheers!!!!!!!!
ALL THE BEST :)
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