find all real solutions to an equation

**absvalue** · September 14th 2009, 07:26 PM

I'm asked to find all real solutions to the equation...

$\sqrt{2}x^{2} - 1 = 0$

Here is what I've done...

$\sqrt{2}x^{2} = 1$

$x^{2} = \frac{1}{\sqrt{2}}$

$x^{2} = \frac{\sqrt{2}}{2}$

$x = \stackrel{+}{-}\sqrt{\frac{\sqrt{2}}{2}}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2}}{\sqrt{2}}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2} \times \sqrt{2}}{2}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2^{3}}}{2}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{8}}{2}$

I'm not feeling confident about this, though. Are these steps correct?

**VonNemo19** · September 14th 2009, 07:35 PM

Originally Posted by absvalue

I'm asked to find all real solutions to the equation...

$\sqrt{2}x^{2} - 1 = 0$

Here is what I've done...

$\sqrt{2}x^{2} = 1$

$x^{2} = \frac{1}{\sqrt{2}}$

$x^{2} = \frac{\sqrt{2}}{2}$

$x = \stackrel{+}{-}\sqrt{\frac{\sqrt{2}}{2}}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2}}{\sqrt{2}}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2} \times \sqrt{2}}{2}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{2^{3}}}{2}$

$x = \stackrel{+}{-}\frac{\sqrt[4]{8}}{2}$

I'm not feeling confident about this, though. Are these steps correct?

They look good to me.

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