What is the solution of the set?

Mathematics

1 answer:

ziro4ka [17]3 years ago

3 0

$x^2=16-4x\ \ \ |+4x\\\\x^2+4x=16\ \ \ \ |+2^2\\\\\underbrace{x^2+2\cdot x\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}=16+2^2\\\\(x+2)^2=20\iff x+2=\pm\sqrt{20}\\\\x+2=-\sqrt{4\cdot5}\ \vee\ x+2=\sqrt{4\cdot5}\ \ \ |-2\\\\x=-2-2\sqrt5\ \vee\ x=-2+2\sqrt5$

Answer: D.

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Agata [3.3K]

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1st Form:

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Step-by-step explanation:

You can see that I put three different forms of answers, but they are all still equivalent answers. The reason why I got three different answers is because I put the inequality in three different calculators and they all got me different answers, but all three are still equivalent. The following three attachments are images of the three calculators that I used to solve.

Hope this helps and answers your question! :)