If \frac{7}{3\sqrt{2} - \sqrt{11}} = a\sqrt{2} + b\sqrt{11}, where a and b are rational nu… — JAMB UTME

If $\frac{7}{3\sqrt{2} - \sqrt{11}} = a\sqrt{2} + b\sqrt{11}$ , where $a$ and $b$ are rational numbers, what is the value of $a + b$ ?

Multiplying both the top and bottom by the conjugate

(3\sqrt{2} + \sqrt{11})

transforms the denominator into

(3\sqrt{2})^2 - (\sqrt{11})^2 = 18 - 11 = 7

. Dividing the numerator

7(3\sqrt{2} + \sqrt{11})

7

simplifies to

3\sqrt{2} + \sqrt{11}

. Matching coefficients yields

a = 3

and

b = 1

, giving

a + b = 4

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

(3\sqrt{2} + \sqrt{11})

\frac{7(3\sqrt{2} + \sqrt{11})}{(3\sqrt{2} - \sqrt{11})(3\sqrt{2} + \sqrt{11})}

Rationalising the denominator eliminates radicals from the bottom of the fraction.

Expand the denominator using the difference of squares formula

(x - y)(x + y) = x^2 - y^2

(3\sqrt{2})^2 - (\sqrt{11})^2 = (9 \times 2) - 11 = 18 - 11 = 7

Squaring each term simplifies the denominator into an integer.

Simplify the overall rational fraction by cancelling common factors.

\frac{7(3\sqrt{2} + \sqrt{11})}{7} = 3\sqrt{2} + \sqrt{11}

The factor of 7 in the numerator and denominator cancels out.

Equate

3\sqrt{2} + 1\sqrt{11}

with

a\sqrt{2} + b\sqrt{11}

to determine the values of

a

and

b

, then sum them.

a = 3, b = 1 \implies a + b = 3 + 1 = 4

Comparing coefficients of corresponding surd terms gives the required values.

Rationalisation of binomial surd denominators using conjugates

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