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Mathematics 21 Online
OpenStudy (anonymous):

x^2+5x-14=0 solve by completing the square

OpenStudy (anonymous):

(5/2)^2

OpenStudy (anonymous):

x^2 + 5x = 14 x^2 +5x +(5/2)^2 = 14 + (5/2)^2 (x + 5/2)^2 = 14 + 25/4 (x + 5/2)^2 = 81/4 taking root of both sides x+5/2 = +/- root(81/4) x = -5/2 +/- 9/2 x = -5+9 or -5-9 ----- ----- 2 2 x = 2 or -7

OpenStudy (anonymous):

\[x^2+5x=14\] \[(x+\frac{5}{2})^2=14+(\frac{5}{2})^2\] \[(x+\frac{5}{2})^2=14+\frac{25}{4}\] \[(x+\frac{5}{2})^2=\frac{81}{4}\] \[x=\frac{5}{2}=\pm \sqrt{\frac{81}{4}}\] \[x+\frac{5}{2}=\pm\frac{9}{2}\] \[x+\frac{5}{2}=\frac{9}{2}\] \[x=\frac{9}{2}-\frac{5}{2}=\frac{4}{2}=2\] or \[x=-\frac{9}{2}-\frac{5}{2}=\frac{-14}{2}=-7\]

OpenStudy (anonymous):

@satellite How r u able to write your answer as above --- i.e with proper fractions etc ???????

OpenStudy (anonymous):

i am writing using latex

OpenStudy (anonymous):

if you come to chat i can show you how. i can't show you here because it will just show up as math

OpenStudy (anonymous):

ok, how do i come to chat??? just click on the chat button???

OpenStudy (anonymous):

yeah just click on chat

OpenStudy (anonymous):

says "group chat"

OpenStudy (anonymous):

will it b possible to hv private chat or not???

OpenStudy (anonymous):

i am there if you want i can show you some basics

OpenStudy (anonymous):

x^2 + 5x = 14 x^2 +5x +(5/2)^2 = 14 + (5/2)^2 (x + 5/2)^2 = 14 + 25/4 (x + 5/2)^2 = 81/4 taking root of both sides x+5/2 = +/- root(81/4) x = -5/2 +/- 9/2 x = -5+9 or -5-9 ----- ----- 2 2 x = 2 or -7

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