SOLUTION: Given a quadratic equation with rational coefficients with a root of 5-i: a)What is another root to the equation?

Question 621340: Given a quadratic equation with rational coefficients with a root of 5-i: a)What is another root to the equation? b) What are the factors of the equation? c) What is the quadratic equation with the given root?
Found 2 solutions by nerdybill, jsmallt9:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Given a quadratic equation with rational coefficients with a root of 5-i:
a)What is another root to the equation?
complex conjugate of 5-i is the other root:
5+i
b) What are the factors of the equation?
s1 = 5-i
s2 = 5+i
(x - (5-i)) and (x - (5+i))
(x-5+i) and (x-5-i)
.
c) What is the quadratic equation with the given root?
b/a = -(s1+s2)
b/a = -(5-i+5+i)
b/a = -(5+5)
b/a = -10
.
c/a = s1s2
c/a = (5-i)(5+i)
c/a = 25-i^2
c/a = 25-(-1)
c/a = 25+1
c/a = 26
.
x^2 + (-10)x + 26
x^2-10x+26

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
If a polynomial (which includes q quadratic) has rational coefficients and a complex root, like 5-i, then the root's conjugate, 5+i, will also be a root.

A quadratic will have two roots. We have found that the two roots for this equation are 5-i and 5+i. Each root will will be in a factor of the form:
(x-r) where the "r" is a root. So factors of this quadratic equation will be:
(x - (5 - i)) and (x - (5 + i))
(Note the parentheses around the roots! These are important.)
which simplify to:
(x - 5 + i) and (x - 5 - i)

A general quadratic equation in factored form is:

where "a" is any number except zero and and are the two roots of the equation. Since there are an infinite number of possible values for "a", there are an infinite number of equations that will have the two roots we have: 5-i and 5+i. So it is impossible to talk about "the" equation having these two roots. I'm assuming that the desired equation is the simplest equation with these two roots. The simplest equation will have an "a" of 1. So our equation is:

or just

The above equation might be acceptable but I'm guessing that you are supposed to multiply out the left side of the equation (so you can see the rational coefficients). The straightforward approach to multiplying (x - 5 + i) and (x - 5 - i) is to multiply each term of one factor times each term of the other and then add like terms. This will work here but it is tedious. A faster approach is to notice that (x - 5 + i)(x - 5 - i) fits the pattern of

with the "a" being "x-5" and "b" being "i". So from the pattern we know that

will be

and to multiply out we can use another pattern, . Using these we get:

which simplifies as follows:

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