Question

solve the quadratic equation 12x^2-3x-9=0 by factoring

Answer 1

1. factor out 3

Answer 2

These questions should be done at wolframalpha to check your answers
http://www.wolframalpha.com/input/?i=12x^2-3x-9%3D0

Answer 3

No, better that you check your answer by multiplication; most people doing factoring problems can use the practice, in my experience. Watching the results of the multiplication come together also give you a better eye for factoring.

Answer 4

12x^2-3x+9=0
>4x^2-x+3=0
>4x^2-4x+3x+3=0
>4x(x-1)+3(x-1)=0
>(x-1)(4x-3)=0
therefore,x-1=0 or,4x-3=0>x=3/4

>x=1

Answer 5

12x^2-3x-9=0
(3)(4x^2-x-3)=0
(3)(x-1)(4x+3)=0
(3x-1)(4x+3)=0
Therefore; x=1/3,-3/4

Answer 6

@Ezack
\[3(x-1)(4x+3) \ne (3x-1)(4x+3)\]Also, no need to incorporate the 3 when solving. \[a(x-1)(4x+3)\]has the same roots for any value of \(a\) that is not 0.

Answer 7

@whpalmer4, thanks for the correction, it's (3x-3) which means x=1, -3/4

Answer 8

@raselazam
\[12x^2-3x-9=0\]\[4x^2-x-3=0\]\[4*-3=-12\]-4,3 are factors of -12 that add to -1
\[(4x^2-4x) +(3x-3) = 0\]\[4x(x-1)+3(x-1) = 0\]\[(4x+3)(x-1) = 0\]\[4x+3=0\]\[4x=-3\]\[x=-\frac{3}{4}\]\[x-1=0\]\[x=1\]So solutions are \[x=1,-\frac{3}{4}\]

Answer 9

you are right actually the answer is 1 and -3/4.but unconsciously I give negative sign ,so the answer 3/4 is wrong.But how can you tell that the process is wrong? how? would you please explain?

Answer 10

Reading your work, you changed some signs...

>4x^2-4x+3x+3=0
>4x(x-1)+3(x-1)=0
>(x-1)(4x-3)=0

in the first line, you've got 3x+3 but it should be 3x-3 (harmless error as you fix it in the next line

but then going from 2nd to 3rd line, you change (4x+3) to (4x-3) which isn't harmless...

Answer 11

When I factor anything, I almost always multiply the factors out to make sure I end up with the polynomial I started with...

\[(x-1)(4x-3) = 4x^2 - 3x -4x -3 = 4x^2 -7x -3\]which isn't what I started with...

Answer 12

Thank you.I found out my mistake.But in this case, is there any way to solve this question applying factoring method?

Answer 13

I thought we did solve this problem by factoring...I don't understand your question.

Answer 14

I just want to know from you is there any alternative way of factorising that is helpful to find out the roots of the equation.