1. ## Radical Equation to standard form trinomial (with math code)

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

Do I square it so that it becomes the following?

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

Thanks.

2. Originally Posted by Kyrie
$\sqrt{2} x^2 + 5x + \sqrt{2} = 0$

Do I square it so that it becomes the following?

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

Thanks.
The first thing you need to understand is that $(A + B + C)^2 \neq A^2 + B^2 + C^2$ so even if the correct approach was to square both sides of the given equation, the result you got is very wrong.

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

3. Originally Posted by mr fantastic
The first thing you need to understand is that $(A + B + C)^2 \neq A^2 + B^2 + C^2$ so even if the correct approach was to square both sides of the given equation, the result you got is very wrong.

$\sqrt{2} x^2 + 5x + \sqrt{2} = 0$ 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.
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
$\frac {-5±\sqrt{25-4(\sqrt{2})(\sqrt{2})}}{2 (\sqrt{2})}$

Step 2 = Evaluate

$\frac {-5±\sqrt{17}}{2 (\sqrt{2})}$

Step 3 = Rationalize denominator

$\frac {-5\sqrt{2}±\sqrt{34}}{2}$

But the answer says the denominator should be 4. What am I doing wrong?

4. Originally Posted by Kyrie
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
$\frac {-5±\sqrt{25-4(\sqrt{2})(\sqrt{2})}}{2 (\sqrt{2})}$

Step 2 = Evaluate

$\frac {-5±\sqrt{17}}{2 (\sqrt{2})}$

Step 3 = Rationalize denominator multiply the numerator and denominator by sqrt(2)

$\frac {-5\sqrt{2}±\sqrt{34}}{2}$

But the answer says the denominator should be 4. What am I doing wrong?

you forgot to multiply the denominator by $\sqrt{2}$ while rationalizing..

$2\sqrt{2} \times \sqrt{2} = 2 \times 2 = 4$

5. Originally Posted by harish21
you forgot to multiply the denominator by $\sqrt{2}$ while rationalizing..

$2\sqrt{2} \times \sqrt{2} = 2 \times 2 = 4$
Ah, thank you, you're right. For some reason I was thinking that the square root just got turned into smoke when I multiplied it in the denominator.

$\frac {-5\sqrt{2}±\sqrt{34}}{4}$