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

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

Answer: Answer

The statement is true because rationalizing the denominator of the algebraic radical expression simplifies directly to

x + \sqrt{x^2 - 1}

Rationalizing the radical expression by multiplying the numerator and denominator by the conjugate

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

leads directly to the simplified form

x + \sqrt{x^2 - 1}

Multiply the numerator and denominator by the conjugate of the denominator,

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

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

Rationalizing the denominator removes radicals from the denominator using the difference of squares identity.

Expand the numerator using

(a+b)^2 = a^2 + 2ab + b^2

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

Simplifying sums of squared radicals and combining like terms.

Expand the denominator using

(a-b)(a+b) = a^2 - b^2

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

Evaluating the difference of squares of square root terms.

Divide the expanded numerator by the simplified denominator.

\frac{2x + 2\sqrt{x^2-1}}{2} = x + \sqrt{x^2-1}

Factoring out 2 from the numerator cancels the denominator of 2.

Rationalizing radical expressions using conjugates and difference of squares.

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