# second grade functionhaving the equations x^2-(m+1)x+m^2=0 and y^2 -(m+3)y+12m+11=0 Between the roots of the 2 equations is established the relation: 2(x1+x2)(y1+y2)=<(x1*x2)(y1*y2) It has to be...

having the equations

x^2-(m+1)x+m^2=0 and y^2 -(m+3)y+12m+11=0

Between the roots of the 2 equations is established the relation:

2(x1+x2)(y1+y2)=<(x1*x2)(y1*y2)

It has to be determined "m" so that the 2 equations to be determined also.

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You need to use the relations between the coefficients and the roots of a polynomial such that:

x1 + x2 = (m+1) ; y1 + y2 = (m+3)

x1*x2 = m^2 ; y1*y2 = 12m + 11

Plugging these relation in the given inequality yields:

2(x1+x2)(y1+y2)<=(x1*x2)(y1*y2)

2 (m+1)(m+3)<=m^2*(12m + 11)

Subtracting m^2*(12m + 11) both sides yields:

2 (m+1)(m+3)-m^2*(12m + 11)<= 0

Opening the brackets yields:

2m^2 + 8m + 6 - 12m^3 - 11m^2 <= 0

- 12m^3 - 9m^2 + 8m + 6 < = 0

12m^3 + 9m^2 - 8m - 6 >= 0

3m^2(4m + 3) - 2(4m + 3) >= 0

(4m+3)(3m^2-2)>=0

Consider both factor positive:

4m+3>=0 => m>=-3/4

3m^2-2>=0 => m=<-sqrt(2/3) or m>=sqrt(2/3)

Consider both factor negative:

4m+3>=0 => m<=-3/4

3m^2-2>=0 => m>=-sqrt(2/3) or m<=sqrt(2/3)

By the relation between coefficients and the roots of the equations,

From the equation,x^2-(m+1)x+m^2=0

x1+x2 = - -(m1+1)/1= m1+m2 ....(1) and

x1*x2 = m^2/1 = m..................(2)

From the equation, y^2 -(m+3)y+12m+11=0,

y1+y2 = m+3 .........................(3)and

y1y2 = 12m+11........................(4) Substituting above values in the given relation, 2(x1+x2))(y1+y2) <= x1x2*y1y2, we get:

2(m+1)(m+3) <= m^2(12m+11). Simplifying we get,

12m^2+9m^2-8m-6 <=0. Factorising, we get:

(4m-3)(m^2-3) < = 0

Therefore, f(m) = 12(m+sqrt(2/3))(m-3/4)(m-sqrt(2/3)) <=0

Therefore, for the values of 3/4 <=m<= sqrt(3/) or x<= - sqrt(2/3) , the inequality and the above relations wll simultaneoudly hold good.

In order to find out the value of the parameter m, we have to resort to the Viete relations, which are the link between the coefficients of an equation and it's roots.

In this case, we have to deal with second grade equations. For the first one, the relations are:

x1+ x2 =- [-(m+1)/1]

x1*x2=m^2

For the second equation, the Viete's relations are:

y1 + y2=-[-(m+3)/1]

y1*y2=12m+11

Now, we'll substitute the found relations into the relation given in the enunciation:

2(x1+x2)(y1+y2)=<(x1*x2)(y1*y2)

2*(m+1)*(m+3)=<(12m+11)*m^2

12m^3 -9m^2 + 8m + 6=<0