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Linear Algebra 18 Online
OpenStudy (anonymous):

Solve x2 − 7x + 12 = 0. x = −3, x = −4 x = 3, x = 4 x = 2, x = 6 x = −2, x = −6

OpenStudy (anonymous):

factor the left part of the equation

OpenStudy (welshfella):

x^2 - 7x + 12 = 0 x^2 -4x - 3x + 12 = 0 now factor the above by grouping

OpenStudy (welshfella):

- that means factor x^2 - 4x and - 3x + 12 then simplify

OpenStudy (anonymous):

I agree

OpenStudy (anonymous):

@DarkMoonZ what did you get after following @welshfella directions

OpenStudy (anonymous):

Ok wait

OpenStudy (anonymous):

x (x - 4) and 3(4-x)

OpenStudy (anonymous):

Sorry I had been AFK lol and thanks @welshfella

OpenStudy (welshfella):

first one is write notice it -3 x + 12 negative 3

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

so its -3(4-x)

OpenStudy (welshfella):

no its -3( x - 4) If you expand that you'll see it comes bakc to -3x + 12 so we got x(x - 4) - 3(x - 4) = 0

OpenStudy (anonymous):

Ok

OpenStudy (welshfella):

now e can simplify this further because there is (x - 4) in both parts

OpenStudy (welshfella):

so take x - 4 out and what have we got left?

OpenStudy (anonymous):

-3x=0

OpenStudy (anonymous):

or just X = -3

OpenStudy (welshfella):

no what i mean by taking x - 4 out is using it as a factor and what left becomes the other factor similar to 3 x + 3y - 3 is common to both 3x and 3y so we take 3 'out' and it becomes a factor 3x + 3y = 3 (x + y) The x+ y becomes the other factor.

OpenStudy (anonymous):

oh ok....

OpenStudy (welshfella):

x(x - 4) - 3(x - 4) = 0 (x - 4) is common so we take it out and whats left ( the x - 3 0 goes in the other parentheses so we have (x - 4)(x - 3) = 0

OpenStudy (welshfella):

now either x - 3 = 0 or x - 4 = 0 making x = 3 , 4

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

thanks

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