Question

in finding the roots of quadratic eq 6x^2-x-2=0 by th method of factorisation,the middle term "-x" can be written as
please answer
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Answer 1

Answer:

$x = \frac{1}{2} \: \: \: \: \: \: \: \: \: \: or\: \: \: \: \: \: \: x = \frac{2}{3}$

Step-by-step explanation:

$6 {x}^{2} - x - 2 = 0 \\ \\ 6 {x}^{2} - 4x + 3x - 2 = 0 \\ \\ 2x(3x - 2) + 1(3x - 2) = 0 \\ \\ (2x - 1)(3x - 2) = 0 \\ \\ x = \frac{1}{2} \: \: \: \: \: \: \: \: \: \: or\: \: \: \: \: \: \: x = \frac{2}{3}$

Answer 2

Step-by-step explanation:

6x^2-x-2=0

6x^2-4x+3x-2=0

2x(3x-2)+1(3x-2)=0

(2x+1)(3x-2)=0

Either 2x+1=0. Or 3x-2=0

x=-1/2 x=2/3

So -x=1/2,-x=-2/3