# How do you solve abs(x^2-2x-3) = sqrt (2- lnx)?

Aug 10, 2016

$x = \left\{2.73282 , 3.21631\right\}$

#### Explanation:

The best way to solve this equation is using an iterative method, after knowing the initial trials. The initial iterations can be found after plotting the curves

${y}_{1} = \left\mid {x}^{2} - 2 x - 3 \right\mid$ and
${y}_{2} = \sqrt{2 - \ln x}$

this plot will cover the interval $0 < x < {e}^{2}$ because in this interval we have feasible solutions.

We now will perform a series of algebraic transformations into the original problem, in order to remove handling difficulties.

First after squaring both equation terms, we consider

${\left\mid {x}^{2} - 2 x - 3 \right\mid}^{2} = 2 - {\log}_{e} x$

and now we call

$f \left(x\right) = {\left({x}^{2} - 2 x - 3\right)}^{2} - 2 + {\log}_{e} x = 0$

Supposing we know an initial guess ${x}_{0}$ we can proceed using the Taylor series first approximation

$f \left({x}_{k + 1}\right) = f \left({x}_{k}\right) + f ' \left({x}_{k}\right) \left({x}_{k + 1} - {x}_{k}\right) + {O}^{2} \left({x}_{k}\right)$

but we need that $f \left({x}_{k + 1}\right) = 0$ then, supposing ${O}^{2} \left({x}_{k}\right)$ small enough, we obtain

${x}_{k + 1} = {x}_{k} - f \frac{{x}_{k}}{f ' \left({x}_{k}\right)}$

From the plot we propose first ${x}_{0} = 2$
and obtain successively
${x}_{1} = 2.66897$
${x}_{2} = 2.72804$
${x}_{3} = 2.73279$
${x}_{4} = 2.73282$

and also ${x}_{0} = 4$
obtaining successively
${x}_{1} = 3.59525$
${x}_{2} = 3.35944$
${x}_{3} = 3.24893$
${x}_{4} = 3.21869$
${x}_{5} = 3.21632$
${x}_{6} = 3.21631$

This way we obtained the equation solution within 6 sd!

Attached a plot with ${y}_{1}$ in red and ${y}_{2}$ in blue, showing the two solutions. 