MHB 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.

When people post problems here in this subforum, they are not seeking help, but rather posting problems for the enjoyment of and challenge to the members of MHB. I do however appreciate the fact that you are trying to help. So, go ahead and have fun with these problems if you like and post full solutions. (Nerd)

[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!