For all x > 4, the expression \frac{x^2 - 4x}{x - 2\sqrt{x}} + \frac{x\sqrt{x} + 8}{\sqrt{… — SAT

For all $x > 4$ , the expression $\frac{x^2 - 4x}{x - 2\sqrt{x}} + \frac{x\sqrt{x} + 8}{\sqrt{x} + 2}$ can be written in the form $ax + b$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Answer: 6

Answer

Factoring the numerator of the first term yields

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

and its denominator yields

\sqrt{x}(\sqrt{x}-2)

. Simplifying this term gives

x + 2\sqrt{x}

. Factoring the numerator of the second term as a sum of cubes gives

(\sqrt{x}+2)(x - 2\sqrt{x} + 4)

, which simplifies to

x - 2\sqrt{x} + 4

. Summing both simplified terms results in

2x + 4

. Matching this to the form

ax+b

gives

a=2

and

b=4

, so

a+b=6

Step-by-Step Solution

Simplify the first term of the expression.

\frac{x^2 - 4x}{x - 2\sqrt{x}} = x + 2\sqrt{x}

Factor

x

from the numerator to get

x(x-4)

and

\sqrt{x}

from the denominator to get

\sqrt{x}(\sqrt{x}-2)

. Rewrite

x-4

as the difference of squares

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

, then cancel the common factor

\sqrt{x}-2

and simplify

\frac{x}{\sqrt{x}}

\sqrt{x}

Simplify the second term of the expression.

\frac{x\sqrt{x} + 8}{\sqrt{x} + 2} = x - 2\sqrt{x} + 4

Recognize

x\sqrt{x} + 8

as a sum of cubes,

(\sqrt{x})^3 + 2^3

. Factor it as

(\sqrt{x}+2)(x - 2\sqrt{x} + 4)

and cancel the common factor of

\sqrt{x}+2

in the denominator.

Add the simplified terms together.

2x + 4

Combine

(x + 2\sqrt{x})

and

(x - 2\sqrt{x} + 4)

by grouping like terms: the

2\sqrt{x}

and

-2\sqrt{x}

cancel out, leaving

2x + 4

Identify the values of

a

and

b

and find

a+b

Comparing

2x + 4

ax + b

gives

a = 2

and

b = 4

. Therefore,

a + b = 2 + 4 = 6

Key Concept

Simplifying rational expressions involving radicals by factoring (difference of squares and sum of cubes).

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic