Question

|2x - 1| = x2

Answer 1

use the quadratic. Do you know what to do know?

Answer 2

nope please explain

Answer 3

Why do you use the quadratic for? Do you know why you have to use it now?

Answer 4

no

Answer 5

Actually no im wrong theres an absolute value.

Answer 6

LOL

Answer 7

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.

Answer 8

Thank you very much, gawww you are so knowledgeable in maths!

Answer 9

You're welcome :)

Answer 10

"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"

Answer 11

i'm confused about that part?

Answer 12

One second...

Answer 13

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.

Answer 14

I didn't apply the derivative theorem for roots properly.

Answer 15

oh i see thanks

Answer 16

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

Answer 17

This website is weird, though.