Mathematics
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OpenStudy (anonymous):
Quadratic Equation:
(x-4)(x-6)=2
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terenzreignz (terenzreignz):
You're supposed to find x, am I right? :)
OpenStudy (anonymous):
haha i dont know :) i think yes?
terenzreignz (terenzreignz):
Well, perform the FOIL method on (x-4)(x-6).
You get...?
OpenStudy (anonymous):
what will you do with 2?
terenzreignz (terenzreignz):
Patience :)
FOIL method on the left-hand-side
you get
x² - 10x + 24
right?
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OpenStudy (anonymous):
yes
terenzreignz (terenzreignz):
so...
x² - 10x + 24 = 2
Now you can more clearly see what to do with the 2, am I right? ;)
OpenStudy (anonymous):
.....
OpenStudy (anonymous):
you cant find x
OpenStudy (anonymous):
then what is it?
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OpenStudy (anonymous):
use FOIL
OpenStudy (anonymous):
know what that is?
OpenStudy (anonymous):
yes. i got x^2-10x+24=2
OpenStudy (anonymous):
then whats next?
terenzreignz (terenzreignz):
Now add -2 to both sides
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OpenStudy (anonymous):
what?
terenzreignz (terenzreignz):
Well, transpose 2, from the right side to the left...
x² - 10x + 22 = 0
OpenStudy (anonymous):
oohh i get it. then what?
terenzreignz (terenzreignz):
Use the quadratic formula. Everything falls into place.
OpenStudy (anonymous):
i really cant. please help me :(
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terenzreignz (terenzreignz):
Given your equation in the form
ax² + bx + c = 0
Your x values are given by
\[\huge \frac{-b \ \pm \ \sqrt{b^2-4ac}}{2a}\]
terenzreignz (terenzreignz):
Plug in and solve.
OpenStudy (anonymous):
what is my a ? my b is 10? my c is 24 ?
OpenStudy (anonymous):
i mean my c is 20?
terenzreignz (terenzreignz):
your b is -10
not just 10
a is 1, as there is no coefficient.
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terenzreignz (terenzreignz):
c is 22, not 24. Look again.
OpenStudy (anonymous):
oh okay :)
then whats next?
OpenStudy (anonymous):
ill just solve it?
terenzreignz (terenzreignz):
Yep.
OpenStudy (anonymous):
i think we're going to do a standard equation not quadratic equation?!
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OpenStudy (anonymous):
standard form rather