Mathematics
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OpenStudy (anonymous):
Hi guys, help me
x^2+3x+2 for this equation how to find the possible x values?
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OpenStudy (anonymous):
First, factorize the equation. Can you do that?
OpenStudy (mathlegend):
Think what numbers when added equal 3 but when multiplied equal 2?
OpenStudy (anonymous):
yes
OpenStudy (mathlegend):
So what numbers when added equal 3 BUT
when multiplied equal 2
OpenStudy (mathlegend):
_+_= 3
_*_= 2
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OpenStudy (mathlegend):
Fill in the blanks
OpenStudy (anonymous):
1,2 is correct?
OpenStudy (mathlegend):
Yes.
OpenStudy (mathlegend):
So now lets break down 3x
x^2+1x+2x+2
OpenStudy (mathlegend):
Now lets factor
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OpenStudy (mathlegend):
x^2+1x then factor 2x+2
OpenStudy (mathlegend):
Are you having trouble @nroprstar
OpenStudy (anonymous):
hi tell me more cleary yar..
OpenStudy (mathlegend):
\[x^{2}+1x+2x+2\]
\[x^{2}+1x \]
2x+2
OpenStudy (mathlegend):
What do we have in common (Letters or Numbers) in the following equation
\[x^{2}\]
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OpenStudy (anonymous):
-1 and -2
OpenStudy (anonymous):
correct!
OpenStudy (mathlegend):
That is not the solution.
OpenStudy (anonymous):
Is not?
OpenStudy (mathlegend):
Okay look...
\[x^{2}+1\]
factors as...
x(x+1)
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OpenStudy (mathlegend):
little typo mean to post: \[x^{2}+1x\]
OpenStudy (anonymous):
No,, I am trying to find eigen values in matrix...
OpenStudy (anonymous):
k mathlegend... tell
OpenStudy (mathlegend):
2x+2
factors as what?
2(x+1)
right
OpenStudy (anonymous):
My mistake, sorry. I probably was thinking of another topic.
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OpenStudy (mathlegend):
So now we have
x(x+1)+2(x+1)
OpenStudy (anonymous):
mm ok
OpenStudy (mathlegend):
All the outside numbers go in one set of parentheses and the inside go in another.
OpenStudy (anonymous):
s
OpenStudy (mathlegend):
(x+2)(x+1)
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OpenStudy (mathlegend):
That is your answer.
OpenStudy (anonymous):
ya now I understood. thanks alot
OpenStudy (mathlegend):
No problem
OpenStudy (anonymous):
so x=-2 or -1
OpenStudy (anonymous):
@MathLegend they are asking for 'possible values of x'
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OpenStudy (anonymous):
if my explanation was correct and we are talking about the same topic, x would be -2 and -1
OpenStudy (mathlegend):
oh, well, okay.
OpenStudy (anonymous):
Hold on. Let me get someone else's opinion.
OpenStudy (anonymous):
Basically MathLegend and I are thinking about the same thing...
OpenStudy (anonymous):
@experimentX
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OpenStudy (experimentx):
you cannot find the value of unknown from an expression.
OpenStudy (experimentx):
you need equations ... f(x) = 0 or a system of equations..
OpenStudy (mathlegend):
Well, being what experimentX said.. then are you sure you were not supposed to just factor @nroprstar
OpenStudy (anonymous):
I was looking of the values of x that would make the equation to zero.
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OpenStudy (anonymous):
Sorry, @nroprstar
OpenStudy (anonymous):
They haven't given a right side so its not an equation, its an expression.
OpenStudy (lgbasallote):
these people are good @pratu043 never doubt them. they are able to *solve* expressions <bows down>