an you please show me how to solve an equation like this?! It's not making sense. x² + 6x = –5
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OpenStudy (abb0t):
Add the 5 and set your polynomial to zero.
OpenStudy (anonymous):
x² + 6x + 5 = 0?
OpenStudy (abb0t):
\[x^2+6x+5 =0\]
notice that it's in the form: \[ax^2+bx+c\]
to factor, you must find TWO numbers whose product is "c", but the sum of the two terms, result in "b". Does that make sense?
It means: n x n = c, but n + n = b
OpenStudy (abb0t):
where the n's are two constants. meaning numbers - obviously.
OpenStudy (anonymous):
Um...still don't get it.
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OpenStudy (anonymous):
you have to find two numbers whose product is 5 and sum is 6
OpenStudy (anonymous):
Oh! Okay!
OpenStudy (abb0t):
Yes.
OpenStudy (anonymous):
But...that's for factoring right? I need to find the value of x...
OpenStudy (anonymous):
yea, you can solve for x by factoring
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OpenStudy (abb0t):
Yes, to find the value(s) of "x". You need to factor.
OpenStudy (anonymous):
or you can use the quadratics formula
OpenStudy (anonymous):
Okay...
OpenStudy (abb0t):
quadratic formula usually makes the algebra more tedious. It's best if you factor.
OpenStudy (anonymous):
Can you guys check my answer once I attempt to figure it out?
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OpenStudy (anonymous):
sure
OpenStudy (abb0t):
Of course.
OpenStudy (anonymous):
Okay, I'm coming up with jibberish. Could you walk me through it?
OpenStudy (anonymous):
make a list of factors of 5
OpenStudy (anonymous):
just 1 and 5
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OpenStudy (anonymous):
and -1 and -5
OpenStudy (anonymous):
yes. so (x + 1)(x + 5) = 0
now you can set those two factors (x + 1 and x + 5) set to 0 and solve for x
OpenStudy (anonymous):
right, but you couldn't use -1 and -5 because those don't add up to 6
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
so solve for x in both
x + 5 = 0 and x + 1 = 0
and they will be your answers
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