1. ## modulus

solve $\displaystyle |x^2-16|=|2x-1|$

2. Square both sides:

$\displaystyle (x^2 - 16)^2 = (2x-1)^2$

$\displaystyle x^4 - 32x^2 + 256 = 4x^2 - 4x + 1$

$\displaystyle x^4 - 36x^2 + 4x + 255 = 0$

You now need to solve this. If you can picture the graphs, you'll know that there is one solution (the most positive) when equating $\displaystyle x^2 - 16 = 2x - 1$

$\displaystyle x^2 - 2x - 15 = 0$

$\displaystyle (x-5)(x+3) = 0$

So, (x-5) and (x+3) are possible roots of the polynomial above. Factoring confirms this.

$\displaystyle x^4 - 36x^2 + 4x + 255 = 0$

$\displaystyle (x-5)(x^3 +5x^2 -11x-51)= 0$

$\displaystyle (x-5)(x^3 +5x^2 -11x-51)= 0$

$\displaystyle (x-5)(x+3)(x^2+2x-17) = 0$

You have a quadratic and two solutions. You can finish this off using the quadratic formula to get the other two remaining roots

3. Hello, Punch!

$\displaystyle \text{Solve: }\;|x^2-16| \:=\:|2x-1|$

Square both sides: .$\displaystyle (x^2-16)^2 \:=\:(2x-1)^2$

. . $\displaystyle \text{We have: }\;\underbrace{(x^2-16)^2 - (2x-1)^2}_{\text{difference of squares}} \:=\:0$

Factor: .$\displaystyle \bigg[(x^2-16) - (2x -1)\bigg]\,\bigg[(x^2-16) + (2x-1)\bigg] \:=\:0$

. . $\displaystyle \text{We have: }\;(x^2-2x-15)(x^2+2x-17) \:=\:0$

$\displaystyle x^2-2x-15 \:=\:0\quad\Rightarrow\quad (x-5)(x+3) \:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:5,\:-3}$

$\displaystyle x^2+2x-17 \:=\:0 \quad\Rightarrow\quad x \:=\:\dfrac{\text{-}2 \pm\sqrt{72}}{2} \quad\Rightarrow\quad \boxed{x \:=\:\text{-}1 \pm 3\sqrt{2}}$

We squared the equation.
. . So we must check for extraneous roots.

But this time, all three roots are acceptable!