WAEC 2009 · Paper 2 · Q1✱✱

(a)
Given that $(\sqrt3 - 5\sqrt2)(\sqrt3 + \sqrt2) = a + b\sqrt6$ , find $a$ and $b$ .
Separate values with commas, e.g. 3, −2
(b)
If $\dfrac{2^{1-y} \times 2^{y-1}}{2^{y+2}} = 8^{2-3y}$ , find $y$ .

Worked solution (try it first)

(a)

Expand the brackets term by term: $\sqrt3 \cdot \sqrt3 + \sqrt3 \cdot \sqrt2 - 5\sqrt2 \cdot \sqrt3 - 5\sqrt2 \cdot \sqrt2$ .
Simplify each term: $3 + \sqrt6 - 5\sqrt6 - 10$ .
Collect like terms: $-7 - 4\sqrt6$ .
Compare with $a + b\sqrt6$ : $a = -7$ and $b = -4$ .

(b)

Add the powers on top (same base 2): $(1 - y) + (y - 1) = 0$ , so the top is $2^0$ .
Divide by subtracting powers: the left side is $2^{0 - (y + 2)} = 2^{-y-2}$ .
Write the right side in base 2: $8^{2-3y} = (2^3)^{2-3y} = 2^{6-9y}$ .
Equate the powers: $-y - 2 = 6 - 9y$ , so $8y = 8$ .
So $y = 1$ .

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