Mathematics
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OpenStudy (anonymous):
Solve for x: (x-2)(x-3)=2
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OpenStudy (anonymous):
Firstly bring it in standard quadratic form:
\[ax^2 + bx + c = 0\]
Can you do it??
OpenStudy (anonymous):
Yes; x^2-2x-3x+6-2=0
OpenStudy (anonymous):
\[x^2-2x-3x+6-2=0\]
OpenStudy (anonymous):
You can add or subtract like terms..
Can you do that??
OpenStudy (anonymous):
\(\large (-2x - 3x) = ?\: ?\)
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OpenStudy (anonymous):
-5x
OpenStudy (anonymous):
So \[x^2-5x+4=0\]
OpenStudy (anonymous):
Can you tell me what is c here by comparing it with the equation I have written at top most..??
OpenStudy (anonymous):
\(c\)
What is c here??
compare it with standard form..
OpenStudy (anonymous):
4 is c
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OpenStudy (anonymous):
Yes can you now tell me the factors of 4??
OpenStudy (anonymous):
1,2,4
OpenStudy (anonymous):
Which factor will give you the sum = 5 and product = 4??
OpenStudy (anonymous):
4 and 1
OpenStudy (anonymous):
For example:
I take 4 = 2*2
is 2 +2 = 5??
No..
So, like wise which factors have sum = 5??
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OpenStudy (anonymous):
Yes you are right till here...
OpenStudy (anonymous):
So would it be (x-4)(x-1)
OpenStudy (anonymous):
so x would be 4 and 1?
OpenStudy (anonymous):
Yes dear you are absolutely right..
OpenStudy (anonymous):
Awesome. Thank you so much for your help.
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OpenStudy (anonymous):
Yes,
\(x = 1 and x = 4\)...
OpenStudy (anonymous):
:)
OpenStudy (anonymous):
Welcome dear...