x^2-(m+1)x+m+4 = 0. To decide the values of m so that the
roots are
negative.
Solution:
We know
that x^2 -(m+1)x + (m+1)^2/4 is a perfect square
{x-(m+1)/2}^2.
Therefore by adding and subtracting
[(m+1)^2]/4, the given expression becomes:
{(x-(m+1)/2)^2 -
(m+1)^2/4 + (m+4) = 0
(x-(m+1)/2)^2 = {(m+1)^2-4(m+4)}/4 =
(m^2-2m-15)/4
Therefore, taking square root, we get the
roots:
x1 = (m+1)/2 + (1/2)sqrt(m^2-2m-15)
Or
X2 = (m+1)/2 -
(1/2)sqrt(m^2-2m-15).
x1 and x2 are real if the
discriminant (m+1)^2-4(m+4) =m^2-2m-15 > 0 Or (m+3)(m-5) > 0. Or for
m<-3 or m >5, the roots are
real...........(R)
Again, satisfying the above
coditions, the both roots are -ve.So
x1<0 and x2
<0 so that x1x2 > 0 or x1x2 = m+4 >
0.
Or m > -4.
Also
x1+x2 = - {-(m+1)} < 0 Or m+1 <0 Or
m
< -1.
So the condtions for m is -4 < m
< -1 .................(N) for both roots to be
negative.
So combining the conditions for reality as at
(R) and negativity as at (N), we should have the condtion for
m:
{ (m<-3) Or (m>5)} & {-4 <m
<-1)} = (-4 < m < -1
)
So if m belongs to the open
interval ]-4, -1 [ , both the roots of x^2-(m+1)x+m+4 = 0
are real and
-ve.
Verification:
m=-3, x^2
-(-3+1)x+(-3+4) = 0 or x^2+2x +1has x = -2+or- sqrt (4-8) .
Satisfying.
m=-5: Equation is x^2+4x-1 = 0, x= -4+or-
sqrt(16+4) ....one root is ppsitive. So the condition does not hold when m
<-4.
m = +6: Equation is x^2 -(6+1)x+ (6+4) = 0. The
roots are : 6+or-sqrt(7^2-4*10) both the roots are positive.
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