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factor the qudratic equation -3X^2-6X+9

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Question by rint
Submitted on 2/17/2004
Related FAQ: [alt.algebra.help] FAQ pointer - read this first
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factor the qudratic equation -3X^2-6X+9


Answer by mathsgirl
Submitted on 4/28/2004
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To factorise a quadratic like this you need to split it into two parts:

1) Multiply the number of x^2 by the constant, i.e. -3*9=-27

2) Find two factors of this number which added together give the number of x, i.e. 3*-9=27 and 3-9=-6

3) Rewrite the quadratic with the x term split up into a sum of these factors, i.e. -3x^2-6x+9=-3x^2+3x-9x+9

4)Factorise the first and second halves separately; -3x^2+3x=3x*(-x)+3x*1=3x(-x+1) and
-9x+9=9*(-x)+9*1=9(-x+1)
so, -3x^2+3x-9x+9=3x(-x+1)+9(-x+1)

5) Factorise again; in this case, there is a common factor of -x+1 so we can write
3x(-x+1)+9(-x+1)=(3x+9)(-x+1)

Hope it helps :-)

 

Answer by boygood
Submitted on 8/6/2004
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the answer of that question is easy!!
a.-3x^2-6x+9
b.-3x^2-6x-9=0
c.(-3x^2-3)+(-3-9)=0
d.x-3=0   x-3=0
e.x=3     x=3


 

Answer by ahsan
Submitted on 11/7/2006
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2x2
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Answer by Narottam
Submitted on 6/26/2007
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1)let a=-3,b=-6 and c=9
2)multiply c*a=9*-3,f(x)=x^2-6x-27
3)completing square=>x^2-6x+9=27+9(adding 9 both sides)
4)=>(x-3)^2=36 => x=3+-6 =>x=9,-3
5)but the original roots will be-->9/a and  -3/a i.e. 3,-1

 

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