Radical Equation to standard form trinomial (with math code)

**Kyrie** · October 11th 2010, 12:45 AM

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

Do I square it so that it becomes the following?

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

Thanks.

**mr fantastic** · October 11th 2010, 12:48 AM

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.

**Kyrie** · October 11th 2010, 03:02 AM

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?

**harish21** · October 11th 2010, 03:41 AM

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..

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

**Kyrie** · October 11th 2010, 06:34 PM

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.

The answer, then, is:

$\frac {-5\sqrt{2}±\sqrt{34}}{4}$

Math Help - Radical Equation to standard form trinomial (with math code)

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Radical Equation to standard form trinomial (with math code)

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