[Quadratics - Discriminant] Help needed (1 Viewer)

[cookiez69]

Show that the roots of the equation

4(m+1)x^2 -4(m-1)x -3 = 0, (m =/= -1)

are real for all real m.

Thanks in advance.

 

[asianese]

 

[braintic]

disc = 16(m-1)^2 - 4 . 4(m+1) . (-3)

=16 [ (m-1)^2 + 3(m+1) ]

=16 [ m^2 - 2m + 1 + 3m + 3 ]

=16 [ m^2 + m + 4 ]

=16 [ (m + 1/2)^2 + 15/4] ... completing square

>0 for all m

so roots are real for all m

 

[braintic]

So

 

[asianese]

Sorry my latex stuffed up. Editted.