For all real numbers x > 1, the expression \sqrt{x + 2\sqrt{x - 1}} - \sqrt{x - 2\sqrt{x -… — GMAT

For all real numbers $x > 1$ , the expression $\sqrt{x + 2\sqrt{x - 1}} - \sqrt{x - 2\sqrt{x - 1}}$ is equal to $2$ .

Cevap: Cevap

The statement is False because the expression equals

2\sqrt{x-1}

(which is less than

2

) whenever

1 < x < 2

The statement asserts that the identity holds for all

x > 1

. However, evaluating

\sqrt{x - 2\sqrt{x-1}}

yields

|\sqrt{x-1} - 1|

. When

1 < x < 2

\sqrt{x-1} < 1

, so

|\sqrt{x-1} - 1| = 1 - \sqrt{x-1}

. Subtraction gives

2\sqrt{x-1} \neq 2

for all

x

(1, 2)

. Thus, the statement is false.

Rewrite the expressions under the outer radicals as perfect squares.

x \pm 2\sqrt{x-1} = (x-1) \pm 2\sqrt{x-1} + 1 = (\sqrt{x-1} \pm 1)^2

Recognize the quadratic pattern

a^2 \pm 2ab + b^2

where

a = \sqrt{x-1}

and

b = 1

Apply the radical identity

\sqrt{u^2} = |u|

to remove the outer square roots.

\sqrt{x + 2\sqrt{x-1}} = |\sqrt{x-1} + 1| = \sqrt{x-1} + 1

and

\sqrt{x - 2\sqrt{x-1}} = |\sqrt{x-1} - 1|

The principal square root of a squared quantity is non-negative, requiring absolute value bars.

Analyze the absolute value

|\sqrt{x-1} - 1|

across the specified domain

x > 1

For

x \ge 2

\sqrt{x-1} \ge 1 \implies |\sqrt{x-1} - 1| = \sqrt{x-1} - 1

. For

1 < x < 2

\sqrt{x-1} < 1 \implies |\sqrt{x-1} - 1| = 1 - \sqrt{x-1}

The sign of the expression inside the absolute value changes at

x = 2

Evaluate the full expression for

1 < x < 2

(\sqrt{x-1} + 1) - (1 - \sqrt{x-1}) = 2\sqrt{x-1}

Since

2\sqrt{x-1} < 2

when

1 < x < 2

, the statement does not hold for all real numbers

x > 1

Nested Radicals and the Principal Square Root Absolute Value Identity

\sqrt{u^2} = |u|

Soru