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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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Sunshine447Best ResponseYou've already chosen the best response.1
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!
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
first factorize the L.H.S. & then gt ur answer
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hi, Sunshine! Would you like to know the whole process from the start?
 one year ago

Sunshine447Best ResponseYou've already chosen the best response.1
yes please! Thanks Spiderman! haha
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.3
Sum = 7 Product = 12 two numbers 4 and 3 x^2 4x  3x + 12 =0 x(x4) 3(x4) (x4) (x3) x = 3 or x = 4
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
All right then. Do you know what the standard form of a quadratic equation is?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.3
@Sunshine447 Did u understand!
 one year ago

Sunshine447Best ResponseYou've already chosen the best response.1
ax + bx + c = 0 right?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\(ax^2 + bx + c = 0\)* What you first do is find the product of \(a\) and \(c\). What are \(a\) and \(c\) here?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.3
@Sunshine447 Did u understand my method????
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Right! What's the product of 1, 12?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Correct! :) Now you have to find two numbers that add to get \(b\) and multiply to get the product of \(a\) and \(c\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Do you know two numbers such that: Their sum is 7. Their product is 12. ?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
You got it! Now, we will 'split' the middle term with these coefficients. \(7x = 3x  4x \Longrightarrow x^2  7x + 12 = x^2  3x  4x + 12\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Do you get it till this point?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
So, \(x^2  3x  4x + 12 = x(x  3)  4(x  3)\). \(x(x  3)  4(x  3) \Longrightarrow ( x  4)( x  3)\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Are you there? Do you get it?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Okay, so \((x  4)(x  3) = 0\). Do you know the 'zeroproduct rule'?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
All right, you have two solutions here. \( \color{Black}{\Rightarrow x  4 = 0}\) \( \color{Black}{\Rightarrow x  3 = 0}\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Can you solve those equations? That is how you get the solutions.
 one year ago

TheViperBest ResponseYou've already chosen the best response.0
\[\LARGE{x^27x+12=0}\]\[\Large{x^23x4x+12=0}\]\[\Large{x(x3)4(x3)=0}\]\[\Large{(x3)(x4)}\]\[\Large{\color{green}{x=3\space or \space4}}\]
 one year ago

TheViperBest ResponseYou've already chosen the best response.0
So A. is the right answer ;)
 one year ago
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