If \frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}} is expressed in the simplified form a + b\sqrt{2}, … — JAMB UTME

If $\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}$ is expressed in the simplified form $a + b\sqrt{2}$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

Answer: 41

Answer

The value of

a + b

is 41.

To rationalise

\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}

, multiply both numerator and denominator by the conjugate

3 + 2\sqrt{2}

. The denominator becomes

3^2 - (2\sqrt{2})^2 = 9 - 8 = 1

. Expanding the numerator gives

(4)(3) + 4(2\sqrt{2}) + 3\sqrt{2}(3) + 3\sqrt{2}(2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

. Comparing with

a + b\sqrt{2}

gives

a = 24

and

b = 17

, so

a + b = 24 + 17 = 41

Step-by-Step Solution

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

The expression becomes

\frac{(4 + 3\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}

Multiplying by the conjugate eliminates surds from the denominator using the difference of two squares identity

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

Simplify the denominator.

(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1

Squaring

2\sqrt{2}

yields

2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8

Expand the numerator.

(4 \times 3) + (4 \times 2\sqrt{2}) + (3\sqrt{2} \times 3) + (3\sqrt{2} \times 2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

Applying the distributive law and grouping rational terms together and like surd terms together.

Identify the values of

a

and

b

and calculate

a + b

a = 24

b = 17

, so

a + b = 24 + 17 = 41

Matching coefficients of the simplified surd form

a + b\sqrt{2}

Key Concept

Rationalisation of binomial surd denominators using conjugates

Estimated Time:2m 0s

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