Find the roots of the following quadratic equation:
2x²-x-1/8=0
Answers
Answered by
3
Hi Friend !!!
Here is ur answer !!!
2x²-x+1/8
= 2x²-x/2-x/2 +1/8
= 2x(x-1/4)-1/2(x-1/4)
= (2x-1/2)(x-1/4)
x = 1/4
Hope it helps u : )
Here is ur answer !!!
2x²-x+1/8
= 2x²-x/2-x/2 +1/8
= 2x(x-1/4)-1/2(x-1/4)
= (2x-1/2)(x-1/4)
x = 1/4
Hope it helps u : )
Anonymous:
there will be +1/8 in the place of -1/8
Yep.. My mistake., Sorry for inconvenience
It should +1/8
it's okk dude : )
I'll edit it
no problem....leave it
Thanks...
: )
Answered by
4
Hey!!!...Here is ur answer
2x^2-x+1/8=0
2x^2-(x/2)-(x/2)+1/8=0
2x (x-1/4)-1/2 (x-1/4)=0
(x-1/4)(2x-1/2)=0
x=1/4 and x=1/4
Hope it will help you
2x^2-x+1/8=0
2x^2-(x/2)-(x/2)+1/8=0
2x (x-1/4)-1/2 (x-1/4)=0
(x-1/4)(2x-1/2)=0
x=1/4 and x=1/4
Hope it will help you
but it's given -1/8
not +1/8
-1/8 is wrong
if we take -1/8 then the answer will not come
yaa
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