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Solve the equation $\left| { x }^{ 2 }+4x+3 \right| = -\left( 2x+5 \right)$.

 Jan 21, 2021

Best Answer 

 #5
avatar+41 
+2

\({x}^{2}+4x+3=-(2x+5)\)

 

\({x}^{2}+4x+3=-2x-5\)

 

\({x}^{2}+4x+3+5=-2x-5+5\)

 

\({x}^{2}+4x+8=-2x \)

 

\(x^2+4x+8+2x=-2x+2x\)

 

\(x^2+6x+8=0\)

 

\(x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}\)

 

\(\sqrt{6^2-4\cdot \:1\cdot \:8}=2 \)

 

\(x_{1,\:2}=\frac{-6\pm \:2}{2\cdot \:1}\)

 

\(x_1=\frac{-6+2}{2\cdot \:1},\:x_2=\frac{-6-2}{2\cdot \:1}\)

 

x=-2

 

x=-4

 Jan 21, 2021
 #1
avatar+41 
+1

What do you want us to solve? Should we solve for x?

 Jan 21, 2021
 #2
avatar+41 
+1

There are 2 solutions to X.

IWantToHelp  Jan 21, 2021
 #3
avatar+41 
+1

X can be -2 or -4

IWantToHelp  Jan 21, 2021
 #5
avatar+41 
+2
Best Answer

\({x}^{2}+4x+3=-(2x+5)\)

 

\({x}^{2}+4x+3=-2x-5\)

 

\({x}^{2}+4x+3+5=-2x-5+5\)

 

\({x}^{2}+4x+8=-2x \)

 

\(x^2+4x+8+2x=-2x+2x\)

 

\(x^2+6x+8=0\)

 

\(x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}\)

 

\(\sqrt{6^2-4\cdot \:1\cdot \:8}=2 \)

 

\(x_{1,\:2}=\frac{-6\pm \:2}{2\cdot \:1}\)

 

\(x_1=\frac{-6+2}{2\cdot \:1},\:x_2=\frac{-6-2}{2\cdot \:1}\)

 

x=-2

 

x=-4

IWantToHelp  Jan 21, 2021
 #4
avatar+267 
+1

@IWantToHelp: There is an edit feature on web2.0calc. If you want to add on to you post just edit it. Instead of posting twice just edit your first post (if you want.)

 Jan 21, 2021
 #6
avatar+41 
+1

Okay, thanks.

IWantToHelp  Jan 21, 2021
 #7
avatar+116124 
0

Excellent, IWTH  !!!!

 

A "Best Answer "   .

 

cool cool cool

CPhill  Jan 21, 2021
 #8
avatar+41 
+1

Thanks @CPhill.

IWantToHelp  Jan 24, 2021

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