ab-x2=(a-b)x=0 need answer plse ans it
Answers
Answered by
1
hey frnd....
ab-x2 = ax-ab = 0
by
ab-2x = 0
ab = 2x
x = ab/2
and from other side...
(a-b)x = 0
a-b = 0
HOPE THIS HELP YOU ☺☺❤❤
ab-x2 = ax-ab = 0
by
ab-2x = 0
ab = 2x
x = ab/2
and from other side...
(a-b)x = 0
a-b = 0
HOPE THIS HELP YOU ☺☺❤❤
shubhamnaik847:
thank you
welcome
hi
hello
any doubt in the question??
yes
plse do it in brief
Answered by
0
hello
lets r&s be the roots then:
r+s=-(b-c)/(a-b)
but r=s:
2r=-(b-c)/(a-b)
r=-(b-c)/[2(a-b)]
also:
r*s=r^2=(c-a)/(a-b)
(b-c)^2/[4 (a-b)^2 ]=(c-a)/(a-b)
(b-c)^2 [4 (a-b)]=(c-a)
b^2-2bc+c^2=4ac-4a^2-4bc+4ab
4a^2-4ab-b^2-4ac+2bc+c^2=0
(2a-b)^2-2c(2a-b)+c^2=0
[(2a-b)-c]^2=0
2a-b-c=0
2a=b+c
lets r&s be the roots then:
r+s=-(b-c)/(a-b)
but r=s:
2r=-(b-c)/(a-b)
r=-(b-c)/[2(a-b)]
also:
r*s=r^2=(c-a)/(a-b)
(b-c)^2/[4 (a-b)^2 ]=(c-a)/(a-b)
(b-c)^2 [4 (a-b)]=(c-a)
b^2-2bc+c^2=4ac-4a^2-4bc+4ab
4a^2-4ab-b^2-4ac+2bc+c^2=0
(2a-b)^2-2c(2a-b)+c^2=0
[(2a-b)-c]^2=0
2a-b-c=0
2a=b+c
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