Prove that the given complex number is a solution of the equation.
\(\displaystyle{x}^{{{2}}}+{4}{x}+{5}={0}\) Write the equation
\((-2-i)^{2}+4(-2-i)+5=0 \)Substitute « with the complex number
\(4+4i+i^{2}+4(-2-i)+5=-0
\)Use square of binomial
\(4+4i+i^{2}-8-4i+5=0\) Use distributive property
\(4+4i+(-1)-8-4i+5=0\) Evaluate the exponent
(4-1-8+5)+(4—4)i Separate real parts and imaginary parts
0+0i=0 Simplify
0=0 Statement is true
Limit
So what I tried is:
From here, using the rule it remains to evaluate
Put the following equation in slope-intercept form: