Answer with Explanation:
[tex]\sqrt{x+2+4}=6\\\\ \sqrt{x+6}=6[/tex]
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
[tex]=\sqrt{30 +6}\\\\=\sqrt{36}=\pm 6[/tex]
One value is 6 and other is ,-6.
So, x=30, is not an extraneous solution.
→→Secondly if your equation is
[tex]\sqrt{x} +2+4=6\\\\ \sqrt{x}+6=6\\\\ \sqrt{x}=6-6\\\\ \sqrt{x}=0\\\\ x=0[/tex]
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
[tex]\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[/tex]
Substituting the value of ,x in original equation
LHS
[tex]=\sqrt{2+2}+4\\\\= \sqrt{4}+4\\\\=2 +4=6[/tex]
=RHS
So, x=2 , is not an Extraneous Solution.
