Answer: x = 9 is not an extraneous solution of the equation.
Step-by-step explanation: [tex]\sqrt{x}= \sqrt{x-8} +2[/tex]
Squaring both the sides of the equation, we get
= [tex](\sqrt{x})^{2} =( \sqrt{x-8} +2)^{2}[/tex]
= [tex]x =( \sqrt{x-8})^{2} +(2)^{2}+ 2(\sqrt{x-8} )(2)[/tex]
= [tex]x= x-8+4+ 2(\sqrt{x-8})(2)[/tex]
Subtracting x on both the sides of the equation, we get
= [tex]-4+ 2(\sqrt{x-8})(2)=0[/tex]
Adding 4 on both the sides of the equation, we get
= [tex]4(\sqrt{x-8})=4[/tex]
Dividing both sides of the equation by 4, we get
= [tex](\sqrt{x-8})=1[/tex]
Squaring again on both the sides of the equation
= [tex]x-8=1[/tex]
= [tex]x=9[/tex]
Check for extraneous solution
Putting the value of x = 9 in the original equation [tex]\sqrt{x}= \sqrt{x-8} +2[/tex], we get
= [tex]\sqrt{9}= \sqrt{9-8} +2[/tex]
= [tex]3= \sqrt{1} +2[/tex]
= [tex]3=3[/tex]
⇒ LHS = RHS
∴ x = 9 is not an extraneous solution of the equation.
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