Question

Find the solution of the square root of the quantity of x plus 2 plus 4 equals 6, and determine if it is an extraneous solution.

Answer 1

Answer with Explanation:

 $\sqrt{x+2+4}=6\\\\ \sqrt{x+6}=6$

Squaring Both sides

→x + 6 =36

→x +6 -6=36-6

→x=30

When we substitute , x=30 in the original equation ,we get

L H S

   $=\sqrt{30 +6}\\\\=\sqrt{36}=\pm 6$

One value is 6 and other is ,-6.

So, x=30, is not an extraneous solution.    

→→Secondly if your equation is

$\sqrt{x} +2+4=6\\\\ \sqrt{x}+6=6\\\\ \sqrt{x}=6-6\\\\ \sqrt{x}=0\\\\ x=0$

Substituting ,the value of x, in original equation

L H S

= 0 +2 +4

=6

=R HS    

So, x=0, is also not an extraneous solution.  

Or, if the equation is

$\rightarrow \sqrt{x+2} +4=6\\\\\rightarrow \sqrt{x+2}=6-4\\\\\rightarrow \sqrt{x+2}=2\\\\ \text{squaring both sides}\\\\ x +2=4\\\\ x=4-2\\\\x=2$

Substituting the value of ,x in original equation

LHS

$=\sqrt{2+2}+4\\\\= \sqrt{4}+4\\\\=2 +4=6$

=RHS

So, x=2 , is not an Extraneous Solution.  

Answer 2

minus 4 both sides

square both sides

x+2=4
minus 2
x=2
 
the answer is 
 x = 2; not extraneous