A student claims that -4i is the only imaginary root of a quadratic polynomial equation that has real coefficients.
a. What is the student’s mistake? (2 points)
b. Write one possible polynomial that has the correct roots from part a in standard form. (3 points)
Thank you!
a. The student's mistake is claiming that -4i is the only imaginary root of the quadratic equation. Since the equation has real coefficients, it is possible for the quadratic equation to have a pair of complex conjugate roots, not just one.
b. One possible polynomial that has the correct roots is:
(x - 4i)(x + 4i) = x² + 16
The polynomial x² + 16 has the roots -4i and 4i, which are complex conjugates. This polynomial has real coefficients.
a. The student's mistake is claiming that -4i is the only imaginary root of a quadratic polynomial equation with real coefficients. In fact, every quadratic equation with real coefficients has two complex roots, whether they are imaginary or real.
b. To find a polynomial that has the correct roots, we have to consider that the complex roots of a polynomial with real coefficients always come in conjugate pairs. The conjugate of -4i is 4i, which means that the correct roots are -4i and 4i.
Using these roots, we can write the polynomial in standard form. The standard form of a quadratic polynomial is given by ax^2 + bx + c, where a, b, and c are real numbers.
Since the quadratic equation has the roots -4i and 4i, it means that the factors of the polynomial are (x - (-4i)) and (x - 4i). Simplifying, we get (x + 4i) and (x - 4i).
To find the polynomial, we multiply these factors together:
(x + 4i)(x - 4i) = x^2 - (4i)^2 = x^2 - 16i^2.
Since i^2 is equal to -1, we can simplify further:
x^2 - 16(-1) = x^2 + 16.
Therefore, one possible polynomial with the correct roots is:
f(x) = x^2 + 16.
a. complex (imaginary) roots occur in pairs
... if -4i is a root , then 4i is also
b. y = x^2 + 16