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 January 5th, 2014, 02:31 PM #1 Newbie   Joined: Jan 2014 Posts: 6 Thanks: 0 How to solve these problems algebraically? 3x^2 + 192 = 0 x^3 + 125 = 0 I don't know where to begin for the first problem. I assumed it would be factoring a sum of squares but 192 is not a square. For the second one I am guessing it is a sum of cubes problem? Second one I got (x + 5)(x^2 - 5x + 25). Would I need to find the zeros and how would I do that? Not sure what it means to "solve" algebraically.
 January 5th, 2014, 02:46 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Re: How to solve these problems algebraically? 1. You can't factor a sum of squares with real numbers. Otherwise, 3x² + 192 = 3(x² + 64) = 3(x - 8i)(x + 8i), hence the roots are 8i and -8i. 2. x³ + 125 = (x + 5)(x² - 5x + 25) = 0 implies x = -5 is a solution. Can you find x if x² - 5x + 25 = 0?
January 5th, 2014, 03:50 PM   #3
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Re: How to solve these problems algebraically?

Hello, GrannySmith!

Quote:
 $3x^2\,+\,192\:=\:0$

$\text{There is only one }x\text{ in the equation.\;\;Solve for it!}$

$3x^2\,+\,192\:=\:0 \;\;\;\Rightarrow\;\;\;3x^2 \:=\:-192 \;\;\;\Rightarrow\;\;\;x^2 \:=\:-64$

$\;\;\;x \:=\:\pm\sqrt{-64} \;\;\;\Rightarrow\;\;\;x \:=\:\pm8i$

Quote:
 $x^3\,+\,125\:=\: 0$

$\text{Same instructions . . .}$

$x^3\,+\,125 \:=\:0 \;\;\;\Rightarrow\;\;\;x^3 \:=\:-125 \;\;\;\Rightarrow\;\;\;x \:=\:\sqrt[3]{-125}$

$\;\;\; x \:=\:-5$

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