## anonymous 5 years ago |2x - 1| = x2

1. anonymous

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

2. anonymous

3. anonymous

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

4. anonymous

no

5. anonymous

Actually no im wrong theres an absolute value.

6. anonymous

LOL

7. anonymous

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.

8. anonymous

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

9. anonymous

You're welcome :)

10. anonymous

"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"

11. anonymous

12. anonymous

One second...

13. anonymous

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.

14. anonymous

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

15. anonymous

oh i see thanks

16. anonymous

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

17. anonymous

This website is weird, though.