0
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ Let's pose $t = \sqrt{y}$ and substitute... $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$−19y2+5y√y+10y+12=0$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.