Answer:
1. Β 1/2 +/-β5
2. Β 1 +/- Β i β6/2.
Step-by-step explanation:
1. Β f(x) = 4x^2 - 8x - 1
x = [-(-8) +/-β((-8)^2 - 4*4*-1)] / (2*4)
x =[ 8 +/- β80] / 8
= 1 +/- 4β5/8
= 1 +/- β5/2
= 1/2 +/-β5
2. By a similar method the zeros of the the second equation are:
x = [-(-4) +/- β((-4)^2 - 4*2*5)] Β / 4
= 4 +/- β(16 - 40) / 4
= 1 +/- β(-24)/4
= 1 +/- i Β β6 * 2 / 4
= 1 +/- Β i β6/2.