why did the take 6alpha-3/k-1 wala equation

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it was taken because the line on the graph cuts on the x axis and not on the y axis 
if it was to be cut at y axis, then the x coordinate would become 0 and thus [ 5alpha +2/ alpha-1 ] would be taken as 0 
but as the line cuts at x axis, the value of y coordinate becomes 0 thus [ 6alpha-3/alpha-1 ] expression is equated as 0 
hence, 6 alpha -3/ alpha - 1 = 0
            6 alpha - 3 = 0 ( alpha - 1)
            6 alpha - 3 = 0 
                 6 alpha = 3 
                    alpha = 3/6
                    alpha = 1/2 
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