- #1

aricho

- 71

- 0

what does

a^2-4a-12

factorise to?

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- Thread starter aricho
- Start date

- #1

aricho

- 71

- 0

what does

a^2-4a-12

factorise to?

- #2

LeonhardEuler

Gold Member

- 860

- 1

- #3

aricho

- 71

- 0

its -6 and 2 isn't it?

gosh I'm a fool sometimes :rofl:

gosh I'm a fool sometimes :rofl:

- #4

LeonhardEuler

Gold Member

- 860

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That's it.

- #5

aricho

- 71

- 0

arrrrrrgggghhhhhh

how about x^2-6x-9

how about x^2-6x-9

- #6

LeonhardEuler

Gold Member

- 860

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- #7

HallsofIvy

Science Advisor

Homework Helper

- 43,021

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For example, to solve a

[tex]\frac{4\pm\sqrt{16-4(1)(-12)}}{2}= \frac{4\pm\sqrt{64}}{2}= \frac{4\pm8}{2}[/tex] so that a= (4+8)/2= 6 and a= (4-8)/2= -2. The factors are (a-6)(a-(-2))= (a-6)(a+2).

More importantly, x

x

I would be inclined to "complete the square": x

[tex]x-3= \pm\3\sqrt{2}[/tex]

so the solutions are [tex]x= 3- 3\sqrt{2}[/tex] and [tex]x= 3+ 3\sqrt{2}[/tex]

[tex]x^2-6x-9= (x-3+3\sqrt{2})(x-3-3\sqrt{2})[/tex].

You could also get that factorization by completing the square originally:

[tex]x^2- 6x- 9= x^2- 6x+ (9-9)- 9= x^2- 6x+ 9- 18= (x-3)^2- 18[/tex]

Now factor as the difference of two squares:

[tex]x^2-6x-9= (x-3)^2-18= ((x-3)+\sqrt{18})((x-3)-\sqrt{18})[/tex]

[tex]= (x-3+3\sqrt{2})(x-3-3\sqrt{3})[/tex]

- #8

sr6622

- 36

- 0

I'm glad no one did the problem for him...

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