# Modulus

• Jun 22nd 2011, 06:19 AM
Punch
Modulus
Solve $4|x-a|=\frac{1}{x-a}$
• Jun 22nd 2011, 06:53 AM
IBstudent
Re: Modulus
hmmm....
say x is 4, and a is 3.5.... substitude them...... 4*0.5=2, 1/0.5=2
as long as the difference is 0.5, its correct.....
• Jun 22nd 2011, 07:11 AM
Prove It
Re: Modulus
Quote:

Originally Posted by Punch
Solve $4|x-a|=\frac{1}{x-a}$

Assuming that $\displaystyle a > 0$

I suggest starting by noting that $\displaystyle |x - a| = \begin{cases}x - a\textrm{ if }x \geq a\\ a - x\textrm{ if }x < a\end{cases}$.

So you need to look at two cases, Case 1:

$\displaystyle x \geq a \implies |x - a| = x - a$, then

\displaystyle \begin{align*}4|x - a| &= \frac{1}{x - a} \\ 4(x - a) &= \frac{1}{x - a} \\ 4(x - a)^2 &= 1 \\ (x - a)^2 &= \frac{1}{4} \\ x - a &= \pm \frac{1}{2} \\ x &= a \pm \frac{1}{2} \\ x &= \frac{2a \pm 1}{2} \end{align*}

Case 2: $\displaystyle x < a \implies |x - a| = a - x$, then

\displaystyle \begin{align*} 4(a - x) &= \frac{1}{x - a} \\ -4(x - a)^2 &= 1 \\ (x - a)^2 &= -\frac{1}{4} \\ x - a &= \pm \frac{i}{2} \\ x &= a \pm \frac{i}{2} \\ x &= \frac{2a \pm i}{2} \end{align*}

So the solutions to the equation $\displaystyle 4|x - a| = \frac{1}{x - a}$ are $\displaystyle x = \left\{\frac{2a - 1}{2}, \frac{2a - i}{2}, \frac{2a + 1}{2}, \frac{2a + i}{2}\right\}$