Bergen
2021-09-24
Answered

Work each problem. Show that -2-i is a solution of the equation ${x}^{2}+4x+5=0$

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oppturf

Answered 2021-09-25
Author has **94** answers

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

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So what I tried is:

$\underset{x\to 0}{lim}\frac{\frac{\mathrm{sin}\left\{x\right\}}{\mathrm{cos}\left\{x\right\}}-\mathrm{sin}\left\{x\right\}}{{x}^{3}}=\underset{x\to 0}{lim}\frac{\left\{\mathrm{sin}\left\{x\right\}\right\}(\frac{1}{\mathrm{cos}x}-1)}{x\cdot {x}^{2}}$

From here, using the rule $\underset{x\to 0}{lim}\frac{\mathrm{sin}\left\{x\right\}}{x}=1$ it remains to evaluate

$\underset{x\to 0}{lim}\frac{\frac{1}{\mathrm{cos}\left\{x\right\}}-1}{{x}^{2}}$

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