solve the following quadratics by factoring x^2=6x x^2=6x-8 6x^2-13x+6=0 5x^2-17x+14=0 12x^2+x=1
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Originally Posted by gumi solve the following quadratics by factoring x^2=6x x^2=6x-8 6x^2-13x+6=0 5x^2-17x+14=0 12x^2+x=1 Why dont you show us some work and we will critique you...for the first one I will give you that $\displaystyle (x-3)^2=9$
for the 2nd one i got (x-3)^2=1 is that right?
Originally Posted by gumi for the 2nd one i got (x-3)^2=1 is that right? Yes that is correct...very good...but it can be simpler factor it into $\displaystyle (x-4)(x-2)$
Hello, For the first one, do you really need to complete the square ? oO It's only asked to factorise ^^ x²=6x x²-6x=0 x(x-6)=0 So...
Originally Posted by Moo Hello, For the first one, do you really need to complete the square ? oO It's only asked to factorise ^^ x²=6x x²-6x=0 x(x-6)=0 So... But on another post of theirs it asked for completing the square so I just took the liberty of assuming that is what she meant
the 3rd one i got (2 + -3x)(3 + -2x) = 0 is that right?
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