The number (1 + i) is a root of the equation x^3-8x^2+14x-12=0. Find the other roots.
a. x = 6
b. x = 7
c. x = 1 + i
d. x = 1 - i
Is it A and D?
yep.
To find the other roots of the equation x^3 - 8x^2 + 14x - 12 = 0, given that (1 + i) is a root, we can use polynomial division or synthetic division. However, since polynomial division can be quite lengthy, let's use synthetic division to solve this problem.
Using synthetic division, we divide the polynomial by the root (1 + i) as follows:
1 + i | 1 - 8 14 - 12
| 1 - 7 7 - 5
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1 - 7 7 -5 + 0
The result of synthetic division gives us the quotient 1, -7, 7, and the remainder 0. The remainder being zero means that (1 + i) is indeed a root of the equation. Now, we are left with a quadratic equation:
x^2 - 7x + 7 = 0
To find the remaining roots, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = -7, and c = 7. Plugging these values into the equation, we get:
x = (7 ± √((-7)^2 - 4 * 1 * 7)) / (2 * 1)
= (7 ± √(49 - 28)) / 2
= (7 ± √21) / 2
So, the remaining roots are:
x = (7 + √21) / 2
x = (7 - √21) / 2
Therefore, the other roots of the equation x^3 - 8x^2 + 14x - 12 = 0 are x = (7 + √21) / 2 and x = (7 - √21) / 2.
In summary, the correct options are:
a. x = 6
d. x = 1 - i