|2x - 1| = x2

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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use the quadratic. Do you know what to do know?
nope please explain
Why do you use the quadratic for? Do you know why you have to use it now?

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no
Actually no im wrong theres an absolute value.
LOL
Going from the definition, you can square both sides again (I'm skipping the sqrt of (2x-1)^2 bit) to get,\[(2x-1)^2=x^4\]You need to solve this equation. When you expand, you'll find,\[x^4-4x^2+4x-1=0\]The rational roots theorem says that if this polynomial is to have a rational root, it should be of the form +/- 1 (in this case). Substituting x=1 shows that it is indeed a root of the polynomial. So we may factor it out. Also, if you take the derivative of the polynomial, you'll find that x=1 is a root of the derivative also. This means x=1 is a double root of your quartic polynomial and you may factor out (x-1)^2. doing this yields\[x^4-4x^2+4x-1=(x-1)^2(x^2+2x-1)\]Setting this to zero leads you to conclude,\[x=1\]is a root (known) and \[x=-1 \pm \sqrt{2}\] from the quadratic.
Thank you very much, gawww you are so knowledgeable in maths!
You're welcome :)
"Also, if you take the derivative of the polynomial, you'll find that x=1 is a root of the derivative also. This means x=1 is a double root of your quartic polynomial and you may factor out (x-1)^2. doing this yields"
i'm confused about that part?
One second...
I was writing something up for that...just scratch it and note that the cubic you obtain when you factor (x-1) out of the quartic also has root x=1 by the rational root theorem. Then you can factor (x-1) from the cubic to obtain the quadratic.
I didn't apply the derivative theorem for roots properly.
oh i see thanks
oops sorry for the triple post, my computer keeps lagging and when i do press it nothing comes up and then after like three tor two posts come up, LOL
This website is weird, though.

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