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A)

In this multi-part problem, we will consider this system of simultaneous equations: 3x+5y6z=2,(i)5xy10yz6xz=41,(ii)xyz=6.(iii)  

Let a=3x, b=5y, and c=6z.  Determine the monic cubic polynomial in terms of a variable t whose roots aret=a, t=b, and t=c. Make sure to enter your answer in terms of tand only t, in expanded form.  (Suggestion: As a first step, you may wish to rewrite the system of equations in terms of a,b, and c.)


 

B)

This is a continuation of the problem above.  Given that (x,y,z) is a solution to the original system of equations, determine all distinct possible values of x+y  (Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple (a,b,c), then determine the possible solutions (x,y,z).)

 Nov 1, 2023
 #1
avatar+1729 
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(A) For any number a, we can substitute a into the cubic equation f(x)=x3+px2+qx+r, and we get: f(a)=a3+pa2+qa+r =a(a2+pa+q)+rThe term (a2+pa+q) is the quadratic expression that has the roots a and −p. We can also think of it as the sum of the roots of the quadratic expression. So if we want a cubic expression with the roots a, b, and c, then the quadratic expression with the roots a and −p is (a+b+c). We can use the same logic for the other two quadratic expressions that we need to create. We get: a(a+b+c)+r=a3+ab+ac+r b(a+b+c)+r=ab2+bb+bc+r c(a+b+c)+r=ac2+bc+cc+rSubtracting the first equation from the second equation, we get: b(a+b+c)+r(a(a+b+c)+r)=0 b(a+b+c)a(a+b+c)=0 (ba)(a+b+c)=0 (ba)(t+5)=0We can do the same for the other pair of equations: (ca)(t+6)=0 (cb)(t+5)=0Multiplying these three equations together, we get the monic cubic expression that we want. (t+5)(t+6)(t+5) =(t2+11t+30)(t+5) =t3+16t2+65t+150Therefore, the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c is: \boxed{t^3+16t^2+65t+150}

 Nov 1, 2023
 #2
avatar+1729 
0

(B)

We can solve the system of equations for x, y, and z by using the following steps:

Multiply the first equation by −6 and the second equation by 5.

Add the two equations together to get −6xy+30yz+30xz=22.

Divide both sides of the equation by −6.

Simplify the equation to get xy−5yz−5xz=−11.

Add the equation xyz=6 to the equation xy−5yz−5xz=−11 to get 6xy−5yz−5xz=−5.

Factor the equation to get (x−1)(6y−5z)=5.

Since x, y, and z are all non-zero, we must have x−1=5 and 6y−5z=−1.

Solve for x and y to get x=6 and y=65z−1​.

Substitute these values into the equation xyz=6 to get 6z=636z−6​.

Solve for z to get z=21​.

Substitute z=21​ into the equation y=65z−1​ to get y=31​.

Therefore, the only possible value of x+y is 6+1/3​ = 19/3.

 Nov 1, 2023

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