Skip to content

# Lars Eighner's Homepage

## LarsWiki

calumus meretrix et gladio innocentis

### Algebra Example of Factoring

{$$\color{red}{ \sqrt{2x+1}=2x -5}$$} {$$\color{green}{-\sqrt{2x+1}=2x -5}$$}

#### Content

{$$\large \sqrt{2x+1} = 2x - 5$$}

#### Solution by factoring.

{\begin{align} \sqrt{2x+1} &= 2x - 5 \cr \sqrt{2x+1} - 2x + 5 &= 0 \cr 2x - \sqrt{2x+1} - 5 &= 0 \cr (2x+1) - \sqrt{2x+1} - 6 &= 0 \cr (\sqrt{2x+1} - 3)(\sqrt{2x+1} +2) &= 0 \end{align}}

The red curve is the given function. The green curve is what the function would be if the sign on the radical on the original function were negative.

##### Case I:

{\begin{align} \sqrt{2x+1} - 3 &= 0 \cr \sqrt{2x+1} &= 3 \cr 2x+1 &= 9 \cr 2x &= 8 \cr x &= 4 \end{align}}

##### Case II:

{\begin{align} \sqrt{2x+1} +2 &= 0 \cr \sqrt{2x+1} &= -2 \end{align}}

So this does not lead to a real root. By squaring both sides, this can be solved:

{\begin{align} \sqrt{2x+1} &= -2 \cr 2x + 1 &= 4 \cr x &= {3 \over 2}\end{align}}

But this is clearly extraneous because {$(0,{3 \over 2})$} is not a point on the original curve.

Algebraically, when this value is substituted back into the original equation, the result is absurd.

{\begin{align} \sqrt{2x+1} &= 2x - 5 \cr \sqrt{2({3 \over 2})+1} &= 2({3\over 2}) - 5 \cr \sqrt{4} &= 2({3\over 2}) - 5 \cr 2 &= - 2 \cr \end{align}}

Graph:

Sources:

Recommended:

Category: Math Algebra

No comments yet.

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

Backlinks

This page is MathAbacus

### August 05, 2017

• HomePage
• WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help