The equation √x+10-6√x+10=5 has:

(B) an extraneous root between -10 and -6

(A) an extraneous root between -5 and -1

(C) a true root between 20 and 25

1953 AMC 12 Problem 23

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23.

The equation

\sqrt{x+10}-\dfrac6{\sqrt{x+10}}=5

has:

an extraneous root between

-5

and

-1

an extraneous root between

-10

and

-6

a true root between

20

and

25

two true roots

two extraneous roots

Multiply by

\sqrt{x+10}

before eliminating the radical

After solving the resulting quadratic, substitute both candidates into the original equation

Multiplying by

\sqrt{x+10}

gives

x+4=5\sqrt{x+10}.

Squaring produces

\begin{gathered} x^2-17x-234\\ =(x-26)(x+9)\\ =0. \end{gathered}

The value

x=26

satisfies the original equation, while

x=-9

makes its left side

1-6=-5,

not

5.

Thus

-9,

which lies between

-10

and

-6,

is extraneous.

The correct answer is B.