Can someone tell me how they got the answer?
Find a polynomial equation with real coefficients that has the given roots.
4i, sqrt5
my answer: x^4-22x^2+80=0
correct answer: x^4+11x^2-80=0
If 4i is a root, there must have been a -4i
and if √5 was a root there must have been a -√5
so the factors were (x-4i)(x+4i)(x+√5)x-√5)
= (x^2 + 16)(x^2-5)
= x^4 - 5x^2 + 16x^2 - 80
= x^4 + 11x^2 - 80
(x+4i)(x-4i)(x-sqrt5)(x+sqrt5)
(x^2 + 16)(x^2-5)
x^4 +11x^2 - 80
I don't know what you did.
What is that saying?
something about "great minds ...." .lol
notice even the posing time was the same
To find a polynomial equation with real coefficients that has the given roots, you can make use of the fact that complex roots occur in conjugate pairs. This means that if 4i is a root, then -4i is also a root.
Given the roots: 4i, -4i, and sqrt(5), let's determine the polynomial equation.
Step 1: Start by writing the quadratic factors for each pair of roots.
- The first pair of roots, 4i and -4i, corresponds to the quadratic factors (x - 4i) and (x + 4i).
- The third root, sqrt(5), corresponds to the linear factor (x - sqrt(5)).
Step 2: Multiply the factors to find the polynomial equation.
(x - 4i)(x + 4i)(x - sqrt(5)) = 0
To simplify the equation, we can multiply the quadratic factors using the difference of squares formula:
(x - 4i)(x + 4i) = (x^2 - (4i)^2) = (x^2 + 16)
Substituting this result back into the polynomial equation:
(x^2 + 16)(x - sqrt(5)) = 0
Expanding further:
x^3 - sqrt(5)x^2 + 16x - 16sqrt(5) = 0
Hence, the correct polynomial equation with the given roots is:
x^3 - sqrt(5)x^2 + 16x - 16sqrt(5) = 0
Note that in your answer, the coefficients for the x^2 term and the constant term are incorrect. The correct polynomial equation is x^3 - sqrt(5)x^2 + 16x - 16sqrt(5) = 0, not x^4 + 11x^2 - 80 = 0.