If #a^2 - 6b^2 - ab = 0#, what is value of #b/a# ?

Question

1571 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-14T10:28:48

# -1/2, or, 1/3#.

Given that, #a^2-6b^2-ab=0#.

#:. 6b^2+ab-a^2=0#.

#:. ul(6b^2+3ab)-ul(2ab-a^2)=0#.

#:. 3b(2b+a)-a(2b+a)=0#.

#:. (2b+a)(3b-a)=0#.

#:. 2b+a=0, or, 3b-a=0#.

#:. 2b=-a, or, 3b=a#.

#:. b/a=-1/2, or, b/a=1/3#.

Answer 2 · 2018-08-14T10:39:19

#b/a=1/3 orb/a=-1/2#

Here ,

#a^2-6b^2-ab=0#

Dividing each term by #a^2!=0#

#1-6b^2/a^2-b/a=0#

#=>6(b/a)^2+b/a-1=0#

For simplicity take #b/a=x#

#:.6x^2+x-1=0#

#=>6x^2+3x-2x-1=0#

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

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

#=>3x-1=0 or 2x+1=0#

#=>x=1/3 or x=-1/2#

Subst. back #x=b/a#

#:.b/a=1/3 orb/a=-1/2#