X^-2 = (2+or-square root of 3) Drawn out please
I showed you already. Please be patient and don't repost. Did you have questions about the solution I presented?
i need to see it worked out to get it
im new at this but is there a way to draw it out to see it?
\[x^{-2}=2 \pm \sqrt3\]\[\frac{1}{x^2}=2 \pm \sqrt3\]\[x^2=\frac1{2 \pm \sqrt3}\]\[x=\sqrt{\frac1{2 \pm \sqrt3}}\]
Just look at my solution one step at a time
If you are confused about a step, ask what was done or why
each line is a single step
\[\large x^{-2} = 2 \pm \sqrt{3}\]\[\large \implies \frac{1}{x^2} = 2\pm \sqrt{3}\]\[\large \implies x^2 = \frac{1}{2\pm \sqrt{3}}\]\[\large \implies x^2 = \frac{1}{2\pm \sqrt{3}} \cdot \frac{2 \mp \sqrt{3}}{2\mp \sqrt{3}}\]\[\large \implies x^2 = \frac{2\pm \sqrt{3}}{4-3} \]\[\large \implies x^2 = \frac{2 \pm \sqrt{3}}{1} \]\[\large \implies x^2 = 2 \pm \sqrt{3}\]\[\large \implies x = \pm \sqrt{2 \pm \sqrt{3}}\]
how did you go from ⟹x2=2±√3/4−3 ⟹x2=2±√3/1
Which line?
how did you go from 5 to 6
4-3 = 1
All I did was simplify the denominator.
oh wow im dumb idk how i missed that
I think the more salient question is the step from 4 to 5
Did you understand what happened there?
yeah i understood that
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