if one root of the equation x^ + 6x+q=0 is twice the other.find the value of 'q'
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let the root be x and y
then,
x= 2y
so,
x+y=-6/1
2y+y= -6
y=-6/3= -2
and x= 2y= -4
therefore,
xy= q (by putting value)
q= 8
then,
x= 2y
so,
x+y=-6/1
2y+y= -6
y=-6/3= -2
and x= 2y= -4
therefore,
xy= q (by putting value)
q= 8
vinudeepu137:
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