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I want to solve the equation $$x^2+(2 x+3) \sqrt{3 x^2+6 x+2}=6 x+5.$$ I tried. Put $t = \sqrt{3 x^2+6 x+2}.$ We have $$x^2+(2 x+3)t - 6x - 5 = 0.$$ Discriminant of this equation (with unknown $x$) is $56-36t+4t^2$. It is not a square number.

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  • $\begingroup$ Do you actually need to put a new variable or just squaring left and right sides will be enough? $\endgroup$ Mar 21, 2013 at 4:29
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    $\begingroup$ I accept all solutions. $\endgroup$ Mar 21, 2013 at 4:31
  • $\begingroup$ You cannot treat $x$ as independent from $t$, since $t^2 = 3 x^2+6 x+2$. So considering it as an equation in $x$ with coefficients involving $t$ is not a very fruitful approach. $\endgroup$ Mar 21, 2013 at 5:37

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Just put it like this:

$(2 x + 3) \sqrt{(3 x^2 + 6 x + 2)} = (6 x + 5 - x^2)$

then square both sides:

$(2 x + 3)^2 (3 x^2 + 6 x + 2) = (6 x + 5 - x^2)^2$

and collect the terms:

$(-1 + x (5 + x)) (7 + x (17 + 11 x))=0$

The answer is straightforward. Will this help?:)

P.S.: the solutions are

$\left\{x\to \frac{1}{22} \left(-17-i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{22} \left(-17+i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{2} \left(-5-\sqrt{29}\right)\right\},\left\{x\to \frac{1}{2} \left(\sqrt{29}-5\right)\right\}$

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  • $\begingroup$ Thank Caran-d'Ache. This help me. $\endgroup$ Mar 21, 2013 at 4:38
  • $\begingroup$ I Think, you use a math soft. Can you solve by hand? $\endgroup$ Mar 21, 2013 at 4:41
  • $\begingroup$ Surely, no math soft :) I used it only to veryfy the solutions (this part is messy): too much to simplify :). But the original solution was obtained without any soft. It is a usual trick. $\endgroup$ Mar 21, 2013 at 4:44

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