$\displaystyle \sqrt{2} x^2 + 5x + \sqrt{2} = 0 $

Do I square it so that it becomes the following?

$\displaystyle 2x^4 + 25x^2 + 2 = 0 $

Thanks.

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- Oct 10th 2010, 11:45 PMKyrieRadical Equation to standard form trinomial (with math code)
$\displaystyle \sqrt{2} x^2 + 5x + \sqrt{2} = 0 $

Do I square it so that it becomes the following?

$\displaystyle 2x^4 + 25x^2 + 2 = 0 $

Thanks. - Oct 10th 2010, 11:48 PMmr fantastic
The first thing you need to understand is that $\displaystyle (A + B + C)^2 \neq A^2 + B^2 + C^2$ so even if the correct approach was to square both sides of the given equation, the result you got is very wrong.

$\displaystyle \sqrt{2} x^2 + 5x + \sqrt{2} = 0 $ can be solved using the usual quadratic formula in the usual way. If you need more help, please show your work and say where you get stuck. - Oct 11th 2010, 02:02 AMKyrie
Thank you for the reply. I guess the radicals scared me. :P

I get stuck when I rationalize the denominator.

Step 1 = Use Quadratic formula

$\displaystyle \frac {-5±\sqrt{25-4(\sqrt{2})(\sqrt{2})}}{2 (\sqrt{2})} $

Step 2 = Evaluate

$\displaystyle \frac {-5±\sqrt{17}}{2 (\sqrt{2})} $

Step 3 = Rationalize denominator

$\displaystyle \frac {-5\sqrt{2}±\sqrt{34}}{2} $

But the answer says the denominator should be 4. What am I doing wrong? - Oct 11th 2010, 02:41 AMharish21
- Oct 11th 2010, 05:34 PMKyrie