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Mathematics 19 Online
OpenStudy (anonymous):

Solve for x: (x-2)(x-3)=2

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):

\(\large (-2x - 3x) = ?\: ?\)

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

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??

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.

OpenStudy (anonymous):

Yes, \(x = 1 and x = 4\)...

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

Welcome dear...

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