Best Answer

2x3 + 2x2 - 1 = 0

Since we have only one change in sign, then there is only one real root, and two imaginary roots.

If we draw the graph of the curve with the given equation in a graphing calculator, and Trace along the curve to the point of intersection of it with the x-axis until the y-value becomes 0, we can find r1 â‰ˆ 0.6.

So that the imaginary roots, r2 and r3 could be:

r1 + r2 + r3 = - an-1/an = - 2/2 = -1

r1r2r3 = - a0/an (odd degree)= - -1/2 = 1/2 = 0.5

So we have:

Let r2 = (c + di), and r3 = c - di, then

r1 + r2 + r3 = -1

0.6 + (c + di) + (c - di) = -1

2c = -1.6

c = -0.8

r1r2r3 = 0.5

0.6(c + di)(c - di) = 0.5

0.6(c2 - d2i2) = 0.5

0.6(c2 + d2) = 0.5

0.6c2 + 0.6d2 = 0.5

0.6d2 = 0.5 - 0.6c2

0.6d2 = 0.5 - 0.6(-0.8)2

0.6d2 = 0.5 - 0.6(0.64)

0.6d2 = 0.116

d2 = 0.116/0.6

d = Â±âˆš(0.116/0.6) â‰ˆ Â±0.4

and

r2 = c + di = -0.8 + 0.4i and r3 = c - di = -0.8 - 0.4i or

r2 = c + di = -0.8 - 0.4i and r3 = c - di = -0.8 + 0.4i

Q: What are the roots of 2x3 plus 2x2 - 1 equals 0?

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