Asked by Vikky

Find a quadratic function with zeros -7i and 7i

Answers

Answered by Steve
(x-7i)(x+7i) = x^2+49
Answered by Vikky
Use the strategy to solve the higher degree polynomial equation and find the roots of the equation.

6x3 + 14x2 - 3x - 7 = 0
Answered by Damon
6 x^3 - 3 x + 14 x^2 - 7 = 0

3 x (2 x^2 - 1) + 7 (2 x^2-1) = 0

(3x+7)( 2x^2-1) = 0

does that help?
Answered by Vikky
If f(x) is a polynomial function of degree 3 and has zeros of
1 and -4i, then find the third zero.
Answered by Steve
complex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?
Answered by Vikky
@ Steve. i still don't get it. should i take the "i" out of the 4? because if i do something like this, (x-)(x+4i) i will have 4ix, of which its confusing.
Answered by Reiny
Vikky, what Steve is telling you that complex roots, as well as irrational roots, always come in conjugate pairs.
e.g. if one root is 5 - 7i, then there has to be a matching 5 + 7i
e.g. if one root is (4 + 3√5)/2 , then there has to be a (4 - 3√5)/2

So, if you know one root is -4i , then there has to be a +4i
Furthermore, since 1 is also a root, we have 3 factors,
(x-1)(x + 4i)(x - 4i)
= (x-1)(x^2 - 16i^2)
= (x-1)(x^2 + 16)
= x^3 - x^2 + 16x - 16

equation:
x^3 - x^2 + 16x- 16 = 0
Answered by Vikky
Now i get it. Thanks a lot
Answered by solomon
find the zeros and multiplicity of the equation, Range and interval q is positive.
q(x)=-x^2(x+1)^2(x-2)

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