Question

Write the rationalized expression of: sqrt9/sqrt3+sqrtx

Answer 1

Answer:

$\frac{3(\sqrt{3} -\sqrt{x})}{3-x}$ or $\frac{3\sqrt{3}-3\sqrt{x} }{3-x}$

Step-by-step explanation:

$\frac{\sqrt{9} }{\sqrt{3} +\sqrt{x} }$

$\frac{3}{\sqrt{3}+\sqrt{x} }$

Multiply by the opposite by using the difference of squares converse formula.

$\frac{3}{\sqrt{3} +\sqrt{x} } *\frac{\sqrt{3} -\sqrt{x}}{\sqrt{3} -\sqrt{x}}$

$\frac{3(\sqrt{3} -\sqrt{x})}{3-x}$

$\frac{3\sqrt{3}-3\sqrt{x} }{3-x}$

Answer 2

Answer:

3 sqrt(3) - 3 sqrt(x)

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3  - x

Step-by-step explanation:

sqrt(9) = 3

so writing the expression as

3

--------------------

sqrt(3) + sqrt(x)

Multiply by the conjugate, sqrt(3) - sqrt(x) in the numerator and denominator

3                            sqrt(3) - sqrt(x)

-------------------- * ------------------

sqrt(3) + sqrt(x)    sqrt(3) - sqrt(x)

Foil the denominator

3 sqrt(3) - 3 sqrt(x)

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sqrt(3) sqrt(3) + sqrt(3x) - sqrt(3x) - sqrt(x^2)

Simplify

3 sqrt(3) - 3 sqrt(x)

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3  - x