[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 said:Solve in real numbers the system of equations
$(3a+b)(a+3b)\sqrt{ab}=14$
$(a+b)(a^2+14ab+b^2)=36$
HallsofIvy said:The first can be "reduced" to a fifth degree polynomial equation and the second to a third degree equation.
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Opalg said:[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]