solve

 

Find all values of x such that \sqrt{4x^2} + \sqrt{x^2} = 6 + 3x.  

 

The square root of 4x² is +2x.  

The square root of x² is +x,  

 

That gives us four propositions to address:  

 

1.    +2x + x  =  6 + 3x    combines to   0 = 6   dismiss as absurdum  

2,    +2x – x  =  6 + 3x    combines to   x = –3  

3.    –2x + x  =  6 + 3x    combines to   x = –1.5  

4.    –2x – x  =  6 + 3x    combines to   x = –1  

_.

Note there is a difference between  x^2 = 4, for which x = ±2,  and  x = √4, for which x = +2 only.

 

√(4x^2) = 2∣x∣     and  √(x^2) = ∣x∣

 

so we have:   3∣x∣ = 6 + 3x

 

Thank you, Alan.  I didn't know about that difference. 

_.