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Mathematics 16 Online
OpenStudy (anonymous):

What are the solutions to the equation x2 + 4x + 5 = 0?

OpenStudy (shadowlegendx):

\[x^2 + 4x + 5 = 0\]

OpenStudy (anonymous):

do you mind complex solutions?

OpenStudy (astrophysics):

Quadratic equation

OpenStudy (shadowlegendx):

Use \[b^2 - 4ac\]

OpenStudy (astrophysics):

\[\huge x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

OpenStudy (anonymous):

@ShadowLegendX nay, 4 is even, complete the square is quicker

OpenStudy (anonymous):

"nah' not ":nay"

OpenStudy (astrophysics):

or complete the square haha

OpenStudy (shadowlegendx):

Well, sometimes formulas are easier for them to remember, especially for tests *shrugs*

OpenStudy (anonymous):

saves you from writing the radical in simplest radical form and then dividing properly is all, less like to make an arithmetic mistake

OpenStudy (anonymous):

@arielle908 the formula is given to you above try it, if it works great, if not we can walk through competing the square

OpenStudy (shadowlegendx):

\[4^2 - 4(1)(5)\] Following the formula \[b^2 - 4ac\] With a = 1 b = 4 c = 5 16 - 20 = - 4 From there we just classify where it falls into

OpenStudy (shadowlegendx):

:)

OpenStudy (anonymous):

\[x^2+4x+5=0\] subtract 5 from both sides, what do you get?

OpenStudy (shadowlegendx):

The answer is not 0 so we know there are two solutions. Also, the answer is negative so we know itʻs complex. Also, -4 is not a perfect square root so we know it is irrational. So we know there are two complex solutions that are irrational in nature. As for the specific solutions, @satellite73 has that part covered :)

OpenStudy (anonymous):

"irrational in nature"??

OpenStudy (shadowlegendx):

two complex conjugates

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