[Quadratics - Discriminant] Help needed (1 Viewer)

cookiez69 · Feb 19, 2013

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 · Feb 19, 2013

braintic · Feb 19, 2013

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 · Feb 19, 2013

asianese said:
$For real roots, $ \Delta\geq0.\\ \Delta &= b^2-4ac\\= [4(m-1)]^2-4[4(m+1)]\times-3 \\ \Rightarrow 16(m-1)^2+48(m+1)\\&=16(m^2-2m+1)+48m+48\\&=16m^2-32m+16+48m+48\\&=16m^2+16m+62\\&=m^2+m+4\\&=\left(m+\frac{1}{2}\right)^2+\frac{15}{4}\geq 0 $ for all real $m.\\ \therefore \Delta \geq0 $ and so the roots of the equation are real.$

So $16m^2+16m+64=m^2+m+4 ??$

asianese · Feb 19, 2013

Sorry my latex stuffed up. Editted.

[Quadratics - Discriminant] Help needed (1 Viewer)

cookiez69

What a stupid name, Nat.

asianese

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braintic

Well-Known Member

braintic

Well-Known Member

asianese

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