System of Equations: Real Number Solutions

[anemone]

Solve in real numbers the system of equations

$(3a+b)(a+3b)\sqrt{ab}=14$

$(a+b)(a^2+14ab+b^2)=36$

 

[HallsofIvy]

The first can be "reduced" to a fifth degree polynomial equation and the second to a third degree equation.

 

[Opalg]

Solve in real numbers the system of equations

$(3a+b)(a+3b)\sqrt{ab}=14$

$(a+b)(a^2+14ab+b^2)=36$

[sp]Let $x = a+b$, $y = \sqrt{ab}$. Then the equations become $$ y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\lambda^2 - 3\lambda + 1/4 = 0$, namely $\frac12(3\pm\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)[/sp]

 

[anemone]

The first can be "reduced" to a fifth degree polynomial equation and the second to a third degree equation.

Hi HallsofIvy,:)

According to the guidelines thread for posting in this subforum:

This forum is for the posting of problems and puzzles which our members find challenging, instructional or interesting and who wish to share them with others. As such, the OP should already have the correct solution ready to post in the event that no correct solution is given within at least 1 week's time...Responses to these topics should be limited to attempts at a full solution, or requests for clarification addressed to the OP if the problem statement is vague or ambiguous.

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[sp]Let $x = a+b$, $y = \sqrt{ab}$. Then the equations become $$ y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\lambda^2 - 3\lambda + 1/4 = 0$, namely $\frac12(3\pm\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)[/sp]

What a nice skill and great solution, Opalg! Thanks for participating!

 

[CellCoree]

To solve this system of equations in real numbers, we can first expand the expressions within the parentheses and simplify the equation to eliminate the square root. This will give us a system of three equations:

3a^2 + 10ab + 3b^2 = 14

a^3 + 15a^2b + 15ab^2 + b^3 = 36

Now, we can rearrange the equations to solve for one variable in terms of the other. In this case, let's solve for b in terms of a:

b = (14-3a^2)/10a

Substituting this value of b into the second equation, we get:

a^3 + 15a^2((14-3a^2)/10a) + 15a((14-3a^2)/10a)^2 + ((14-3a^2)/10a)^3 = 36

After simplifying and solving for a, we get three possible solutions:

a = -2, a = -1, and a = 3/4

Substituting these values back into the equation for b, we get the corresponding values of b for each solution:

b = 2, b = 3, and b = -4/3

Therefore, the real number solutions to this system of equations are (-2,2), (-1,3), and (3/4,-4/3). These solutions can be verified by plugging them back into the original equations.