What is the simplified form of \frac{6}{3 + \sqrt{3}} after rationalizing the denominator? — JAMB UTME

What is the simplified form of $\frac{6}{3 + \sqrt{3}}$ after rationalizing the denominator?

3 - \sqrt{3}

Multiplying the numerator and denominator by the conjugate

3 - \sqrt{3}

converts the denominator into

3^2 - (\sqrt{3})^2 = 6

. Dividing

6(3 - \sqrt{3})

6

yields

3 - \sqrt{3}

Identify the conjugate of the denominator

The conjugate of

3 + \sqrt{3}

3 - \sqrt{3}

To eliminate the surd from the denominator, multiply by its conjugate.

Multiply the numerator and denominator by the conjugate

\frac{6(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}

Multiplying by

\frac{3 - \sqrt{3}}{3 - \sqrt{3}}

is equivalent to multiplying by 1.

Simplify the denominator using difference of two squares

(3)^2 - (\sqrt{3})^2 = 9 - 3 = 6

The product

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

simplifies the denominator to a rational number.

Divide the numerator by the denominator

\frac{6(3 - \sqrt{3})}{6} = 3 - \sqrt{3}

Canceling out the common factor of 6 gives the final simplified surd expression.

Rationalization of Binomial Denominators

Soru