Solve x^4 - 12x - 5 = 0 given that -1 + 2i is a root.
I’ve used synthetic division to get the answers x = 1 +/- 2i , 1 +/- sqrt 2
I’ve having trouble checking the answers though because of that imaginary number. How can I check these kind of problems?
If (-1+2i) is a solution, its complex conjugate (-1-2i) must be a solution.
In other words
x = -1+2i
x = -1-2i
are solutions
so
x+1-2iand x+1+2i are both factors
so their product should be a factor
I get x^2+2x+5
now do your division
I get
x^2-2x-1
well
that is
(x-1)(x-1)
so my four solutions are those two complex conjugates and x=1 and x =1 again
Whoa - a bit hasty there
x^2-2x-1
can not factor that so easily!
I will have to solve it
x^2-2x-1 = 0
x = (1/2)(2 +/-sqrt(4+4))
=(1/2)(2 +/-2sqrt(2))
x = 1+sqrt(2)
or
x = 1-sqrt(2)
are the last two solutions
Solve x^4 - 12x - 5 = 0 given that -1 + 2i is a root.
Well, checking solutions in polynomials is kinda like checking yourself before you wreck yourself. But since you're having trouble with imaginary numbers, here's a not-so-imaginary way to check them!
To check if -1 + 2i is a root of the equation, simply substitute it back into the equation and see if it holds true. So let's do that:
(-1 + 2i)^4 - 12(-1 + 2i) - 5 = 0
Now, let the clown show begin! Expand those powers!
(1 - 4i + 4i^2 - 8i^3 + 16i^4) + 12 - 24i - 5 = 0
Now, i^2 is defined as -1, and i^3 is defined as -i, and i^4 is defined as 1. So let's simplify further:
(1 - 4i - 4 + 8i + 16) - 24i + 12 - 5 = 0
Phew, almost there! Let's combine like terms:
25 - 20i - 24i + 7 = 0
Now, add up those terms:
32 - 44i = 0
Uh-oh! As we can see, the left side is not equal to zero. So unfortunately, -1 + 2i is not a root of the equation. Time to go back to the drawing board and double-check your calculations.
To check the roots of the equation, you need to substitute each value into the equation and see if the left side of the equation equals zero.
Let's start by checking x = 1 + 2i.
Substitute x = 1 + 2i into the equation:
(1 + 2i)^4 - 12(1 + 2i) - 5
Now, expand (1 + 2i)^4. This can be done by using the binomial theorem or by multiplying it out manually:
(1 + 2i)^4 = (1 + 2i)(1 + 2i)(1 + 2i)(1 + 2i)
= (1 + 2i)(1 + 4i + 4i^2)(1 + 4i + 4i^2)
= (1 + 2i)(1 + 4i + 4(-1))(1 + 4i + 4(-1))
= (1 + 2i)(1 + 4i - 4)(1 + 4i - 4)
= (1 + 2i)(-3 + 4i)(-3 + 4i)
= (-3 - 14i + 8i^2)(-3 + 4i)
= (-3 - 14i + 8(-1))(-3 + 4i)
= (-3 - 14i - 8)(-3 + 4i)
= (-11 - 14i)(-3 + 4i)
= 33 - 11i + 44i - 56
= -23 + 33i
Now substitute this value back into the original equation:
(-23 + 33i) - 12(1 + 2i) - 5
Simplify further:
-23 + 33i - 12 - 24i - 5
= -40 + 9i
Since the left side of the equation (-40 + 9i) does not equal zero, x = 1 + 2i is not a valid root.
Similarly, you can check the other roots, x = 1 - 2i, x = 1 + sqrt(2), and x = 1 - sqrt(2), by substituting them into the equation and checking if the left side equals zero.