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I have this equation:

$-19y^2 + 5y\sqrt{y} + 10y + 12 = 0$

I'm stuck with finding $y$. I tried numerous adjustments like this for example:

$y*(-19y+5\sqrt{y}+10)+12 =0$

but it didn't get me any closer to a solution. I know it must be really simple but right now I'm stuck for hours. Can you give me a hint how to solve this equation?

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    $\begingroup$ Let's pose $t = \sqrt{y}$ and substitute... $\endgroup$
    – the_candyman
    Jun 23, 2014 at 17:22
  • $\begingroup$ Set $x = \sqrt y$. Then it will be a quartic. I notice that one root is $x = 1$, so you can factor out $(x-1)$ and get a cubic. I haven't checked after that to see if there are other simplifications. $\endgroup$
    – Stephen Montgomery-Smith
    Jun 23, 2014 at 17:23
  • $\begingroup$ @StephenMontgomery-Smith $1=x=\sqrt{y}$ is not a root of the expression. $\endgroup$
    – mike
    Jun 23, 2014 at 17:34
  • $\begingroup$ @mike Oops, you are correct. I added the coefficients incorrectly. $\endgroup$
    – Stephen Montgomery-Smith
    Jun 23, 2014 at 17:35
  • $\begingroup$ Numerical solution showed that there is only one positive solution: $y=1.30971$. You can get closed-from solution from $-19t^4 + 5t^3 + 10t^2 + 12 = 0$ using the formulas at wikipedia.com $\endgroup$
    – mike
    Jun 23, 2014 at 17:46

2 Answers 2

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Is this a quadratic equation?

No, because for that to be the case, all exponents would have to be natural numbers. But $\sqrt y=y^\frac12$ and $\frac12\not\in$ N. However, by letting $x=\sqrt y$, you'll have a quartic equation in x.

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$−19y2+5y√y+10y+12=0$

Assuming

$\sqrt y=t$ then, $y=t^2,y^2=t^4$

So, it becomes $-19t^4+5t^3+10t^2+12=0$

You see that it is missing expression with $t$, so that will make this difficult to calculate. Alternatively, I would suggest throwing the whole expression on a graph and find the x-intersects.

you will get a value of $t\approx 1.14$ , so your $y$ will become 1.299

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