Find a quadratic function with zeros -7i and 7i
(x-7i)(x+7i) = x^2+49
Use the strategy to solve the higher degree polynomial equation and find the roots of the equation.
6x3 + 14x2 - 3x - 7 = 0
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?
If f(x) is a polynomial function of degree 3 and has zeros of
1 and -4i, then find the third zero.
complex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?
@ 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.
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
Now i get it. Thanks a lot
find the zeros and multiplicity of the equation, Range and interval q is positive.
q(x)=-x^2(x+1)^2(x-2)
To find a quadratic function with zeros -7i and 7i, we can use the fact that complex zeros come in conjugate pairs for quadratic functions with real coefficients.
The conjugate of -7i is 7i. Therefore, the zeros of the quadratic function are -7i and 7i.
Since complex zeros come in conjugate pairs, we can write the quadratic function in factored form as:
(x - (-7i))(x - 7i) = 0
Simplifying this expression:
(x + 7i)(x - 7i) = 0
Next, we can apply the difference of squares formula to expand the factored form:
(x + 7i)(x - 7i) = x^2 - (7i)^2 = x^2 - 49(i^2)
Remember that i^2 = -1:
x^2 - 49(-1) = x^2 + 49
So, the quadratic function with zeros -7i and 7i is:
f(x) = x^2 + 49