a) let P(x)= 2x^2+x-2=0
P(0)= 0+0-2 = -2 which is negative.
P(1)= 2+1-2 = 1 which is positive.
Since there is a switch in sign, curve is continuous. Therefore, there is a root between 0 and 1.
b) P(0.5)= 2(0.25)+0.5-2= -1 which is negative.
From a), P(1) is positive. Therefore root is between 0.5 and 1.
P(0.75)= 2(0.75)^2+0.75-2=-0.125 which is negative.
Since P(0.75) is closer than P(1) to 0, the root is approximately 0.75.
