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


Read or Post Comments

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

HOME

The best way to look for anything in LarsWiki is to use the search bar.

Page List

Categories

Physics Pages

Math Pages

Math Exercises

Math Tools

Sections