Question

How to solve this absolute value equation. Only users sure of this reply

Answer 1

\[\left| x ^{2}-4x+3 \right|+2x+5=0\]

Answer 2

And answer is
\[\huge \left\{ -4, -1-\sqrt{3} \right\}\]

Answer 3

\(\normalsize\color{blue}{ │x^2-4x+3│+2x+5=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ │x^2-4x+3│=-2x-5\LARGE\color{white}{ \rm │ }}\)

then you get 2 possibilities.

possibility 1, \(\normalsize\color{blue}{ x^2-4x+3=-2x-5\LARGE\color{white}{ \rm │ }}\)
possibility 2, \(\normalsize\color{blue}{ x^2-4x+3=+2x+5\LARGE\color{white}{ \rm │ }}\)

Answer 4

set builder notation :D I was never a fan of it though...

Answer 5

Do you need more assistance solving it ?

Answer 6

I have gone mad thank you lol

Answer 7

my brain is not working

Answer 8

:)

Answer 9

1) \(\normalsize\color{blue}{ x^2-4x+3+2x+5=0\LARGE\color{white}{ \rm │ }}\)
2) \(\normalsize\color{blue}{ x^2-4x+3+2x+5=0\LARGE\color{white}{ \rm │ }}\)

Answer 10

1) \(\normalsize\color{blue}{ x^2-4x+3+2x+5=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x^2-2x+8=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x^2-2x+1=-7\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (x-1)^2=-7\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (x-1)=±i\sqrt{7}\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x=1±i\sqrt{7}\LARGE\color{white}{ \rm │ }}\)

Answer 11

2) \(\normalsize\color{blue}{ x^2-4x+3-2x-5=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x^2-6x-2=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x^2-6x-2+11=0+11\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x^2-6x+9=11\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (x-3)^2=11\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x-3=±\sqrt{11}\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x=3±\sqrt{11}\LARGE\color{white}{ \rm │ }}\)

Answer 12

This is it, and there are 4 solutions (because you have quadratic, this is 3 already, and PLUS the absolute value.)

Answer 13

this is *2 already

Answer 14

thank you but i got it