# Determine the set of values of m for which both roots of equation x^2 - (m+1)x + m+4 = 0 are real and negative.

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### 3 Answers

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

If the equation has ral roots, then the discriminant must be positibs:

Let us verify:

d = b^2 - 4ac > 0

==> (m+1)^2 - 4*1* (m+4) > 0

==> m^2 + 2m + 1 - 4m - 16 > 0

==> m^2 -2m - 15 > 0

==> (m-5)(m+3) >0

==> m-5 > 0 and (m+3)>0 OR (m-5)< 0 and m+3 < 0

==> m > 5 and m > -3 OR m < 5 and m < -3

==> m belongs to ( 5, inf) U (-inf, -3)...........(1)

Now in order for the roots to be negative then :

x1+ x2 < 0 and x1*x2 > 0

But x1 + x2 = -b/a = (m+1) < 0 ==> m < -1`

Also, x1*x2 = c/a = (m+4) > 0 ==> m > -4

==> m belongs to ( -inf, -1) U (-4, inf)...........(2)

From (1) and (2) we conclude:

m belongs to (-inf, -3) U (5, inf )

The first constraint is the equation has 2 real roots, that means that the discriminant of the equation, delta, is positive:

delta>0

delta = b^2 - 4ac

Let's identify the coefficients:

a = 1, b = (m+1) , c = m+4

putting the values of coefficients in the formula of delta, we'll get:

delta = (m+1)^2 - 4(m+4)

delta = m^2 + 2m + 1 - 4m - 16 > 0

Combining like terms, we'll get:

m^2 - 2m - 15>0

We'll apply the quadratic formula:

m1 = [2 + sqrt(4+60)]/2

m1 = (2+8)/2

m1 = 5

m2 = (2-8)/2

m2 = -3

Delta is positive on the interval (-inf., -3) or (5,+inf.).

There are 2 more constraints for this issue:

- the sum of the roots has to be negative, because the roots are negative;

- the product of the roots has to be positive, because the roots are both negative.

We'll translate the constraints into formula:

S = x1 + x2 = -b/a = m+1<0

m<-1

P = x1*x2 = c/a = m+4>0

m>-4

The common intervals of values, that are satisfying all the constraints, are:

**(-inf., -3) or (5,+inf.)**

**m<-3 or m>5**

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.