A quadratic equation with real coefficients and leading coefficient 1, has x = -bi as a root. Write the equation in general form.

Question

mathmate · Answer

A polynomial with real coefficients has complex roots in conjugate pairs.
Hence if one root is -bi, the conjugate root is +bi.  The quadratic is then
f(x)=a(x+bi)(x-bi).
If the leading coefficient (coefficient of the term of highest degree, namely x^2) is one, then a=1, giving
f(x)=(x+bi)(x-bi),
Expand the expression and write in general form to complete the solution.

anonymous · Answer

how will i write in in general form? :)

mathmate · Answer

See: https://www.mathsisfun.com/algebra/polynomials-general-form.html

anonymous · Answer

\[x ^{2}-bi^2 =0 ???

mathmate · Answer

Almost!  But do not forget $ab^2$ does not equal $(ab)^2=a^2b^2$, and that $i^2=-1$

$x ^{2}-(bi)^2 =0$
Expand the second term and simplify to get the general form.