Correct option is (C) an extraneous root between -10 and -6

Given equation is

\(\sqrt{x+10}-\frac{6}{\sqrt{x+10}}=5\)      _____________(1)

\(\Rightarrow\) x + 10 - 6 \(=5\sqrt{x+10}\)

\(\Rightarrow\) x+4 \(=5\sqrt{x+10}\)

\(\Rightarrow(x+4)^2=25(x+10)\)      (By squaring both sides)

\(\Rightarrow x^2+8x+16=25x+250\)

\(\Rightarrow x^2-17x-234=0\)

\(\Rightarrow x^2-26x+9x-234=0\)

\(\Rightarrow x(x-26)+9(x-26)=0\)

\(\Rightarrow(x-26)(x+9)=0\)

\(\Rightarrow\) x - 26 = 0 or x+9 = 0

\(\Rightarrow\) x = 26 or x = -9

Hence, roots of transformed equation are -9 and 26.

Put x = 26 in equation (1), we obtain

\(\sqrt{26+10}-\frac{6}{\sqrt{26+10}}=5\)

\(\Rightarrow\) \(\sqrt{36}-\frac{6}{\sqrt{36}}=5\)

\(\Rightarrow6-\frac66=5\)

\(\Rightarrow\) 6 - 1 = 5

\(\Rightarrow\) 5 = 5    (Satisfied)

Hence, x = 26 is a root of given equation.

Put x = -9 in equation (1), we obtain

\(\sqrt{-9+10}-\frac{6}{\sqrt{-9+10}}=5\)

\(\Rightarrow\) \(\sqrt{1}-\frac{6}{\sqrt{1}}=5\)

\(\Rightarrow\) 1 - 6 = 5

\(\Rightarrow\) - 5 = 5    (Not satisfied)

\(\therefore\) x = -9 is not a root of given equation but x = -9 is a root of its transformed equation.

\(\therefore\) x = -9 is an extraneous root of given which lies between -10 and -6.    \((\because-10<-9<-6)\)

Hence, an extraneous root of given equation lies between -10 and -6.

Correct option is C) an extraneous root between -10 and -6