Sunshine447
Solve the equation by factoring:
x2 - 7x + 12 = 0
A)
x = 3 or 4
B)
x = -3 or 4
C)
x = 3 or -4
D)
x = -3 or -4
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Sunshine447
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please don't just give me the answer.... I want to actually try to learn how to do it. Also, know how to do this the "plug and play" method, but I need to know how to do it when it's not multiple choice. Thanks!
jiteshmeghwal9
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sorry dude
Sunshine447
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why?
jiteshmeghwal9
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first factorize the L.H.S. & then gt ur answer
Sunshine447
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LHS?
ParthKohli
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Hi, Sunshine! Would you like to know the whole process from the start?
Sunshine447
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yes please! Thanks Spiderman! haha
Yahoo!
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Sum = -7
Product = 12
two numbers -4 and -3
x^2 -4x - 3x + 12 =0
x(x-4) -3(x-4)
(x-4) (x-3)
x = 3 or x = 4
ParthKohli
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All right then.
Do you know what the standard form of a quadratic equation is?
Yahoo!
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@Sunshine447 Did u understand!
Sunshine447
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@ParthKohli yes
ParthKohli
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Okay, what is it?
Sunshine447
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ax + bx + c = 0 right?
ParthKohli
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\(ax^2 + bx + c = 0\)*
What you first do is find the product of \(a\) and \(c\).
What are \(a\) and \(c\) here?
Yahoo!
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@Sunshine447 Did u understand my method????
Sunshine447
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a=1 and c=12
ParthKohli
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Right! What's the product of 1, 12?
Sunshine447
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1
12
ParthKohli
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Correct! :)
Now you have to find two numbers that add to get \(b\) and multiply to get the product of \(a\) and \(c\).
ParthKohli
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Do you know two numbers such that:
Their sum is -7.
Their product is 12.
?
Sunshine447
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no
ParthKohli
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Hmm, you may find them by trial and error.
Note that a negative number times a negative number is a positive number, so those numbers must be negative if the sum of negative.
Sunshine447
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-3 and -4
ParthKohli
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You got it!
Now, we will 'split' the middle term with these coefficients.
\(-7x = -3x - 4x \Longrightarrow x^2 - 7x + 12 = x^2 - 3x - 4x + 12\)
ParthKohli
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Do you get it till this point?
Sunshine447
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yeah i think so
ParthKohli
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Okay, now we basically factor by 'grouping'.
What you must do is factor the first two and last two terms.
\(x^2 - 3x = x(x - 3)\)
\(-4x + 12 = -4(x - 3)\)
Get it till here?
Sunshine447
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yeah
ParthKohli
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So, \(x^2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3)\).
\(x(x - 3) - 4(x - 3) \Longrightarrow ( x - 4)( x - 3)\)
ParthKohli
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Are you there? Do you get it?
Sunshine447
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yeah
ParthKohli
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Okay, so \((x - 4)(x - 3) = 0\).
Do you know the 'zero-product rule'?
Sunshine447
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i dont think so
ParthKohli
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All right, you have two solutions here.
\( \color{Black}{\Rightarrow x - 4 = 0}\)
\( \color{Black}{\Rightarrow x - 3 = 0}\)
Sunshine447
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okay...
ParthKohli
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Can you solve those equations?
That is how you get the solutions.
Sunshine447
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x=4 and 3
Sunshine447
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Thanks!
ParthKohli
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You're welcome!
TheViper
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\[\LARGE{x^2-7x+12=0}\]\[\Large{x^2-3x-4x+12=0}\]\[\Large{x(x-3)-4(x-3)=0}\]\[\Large{(x-3)(x-4)}\]\[\Large{\color{green}{x=3\space or \space4}}\]
TheViper
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So A. is the right answer ;)
TheViper
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Understood ???