2x^4+3=12 but x must be a decimal

Question

tkhunny · Answer

Do you have a calculator with a square root button?

anonymous · Answer

ya

tkhunny · Answer

That's about all you need.

2x^4 + 3 = 12

Subtract 3

2x^4 = 9

Divide by 2

x^4 = 9/2

$x^{2} = \pm 3/\sqrt{2}$

Now what?

anonymous · Answer

I sqrt $$\pm3\sqrt{2}$$

anonymous · Answer

Wait no u cant have a sqrt as a denominator so i have to mutiply both the numerator and denominator by sqrt2

tkhunny · Answer

I don't normally care about that, but we're not done!  That's only x^2.  You need 'x'.

I'm assuming we're staing in the Real Numbers. If we have x^2 that CANNOT be negative, so we're down to $x^{2} = \dfrac{3}{\sqrt{2}}$.  Can you find x?

anonymous · Answer

is it $$\pm1.46$$ because i have to round to two decimal places

tkhunny · Answer

You keep making large jumps.  Try to figure out how to go one step at a time.

We have $x^{2} = \dfrac{3}{\sqrt{2}}$.  This gives $x = \pm\dfrac{\sqrt{3}}{\sqrt[4]{2}}$.

tkhunny · Answer

This may just be me, but I wuold suggest ALWAYS giving an exact answer and supply the less exact answer if required.

anonymous · Answer

okay