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
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?
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.
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
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)
q(x)=-x^2(x+1)^2(x-2)
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