Test Index

WAEC Mathematics 2024 Paper

Question : 6

Total: 50

If (3 - 4

\sqrt{2}

\sqrt{2}

\sqrt{2}

, find the value of b

Solution:

Concept:

Expanding two binomials involving surds and then comparing the rational and irrational parts.

Explanation:

Given

(3 - 4\sqrt{2})(1 + 3\sqrt{2}) = a + b\sqrt{2}

First expand the left-hand side:

(3)(1) + (3)(3\sqrt{2}) + (-4\sqrt{2})(1) + (-4\sqrt{2})(3\sqrt{2})

This becomes:

3 + 9\sqrt{2} - 4\sqrt{2} - 12(\sqrt{2} \times \sqrt{2})

Since

\sqrt{2}\times \sqrt{2} = 2

, we get:

3 + 9\sqrt{2} - 4\sqrt{2} - 24

Combine the constants and like surds:

(3 - 24) + (9\sqrt{2} - 4\sqrt{2}) = -21 + 5\sqrt{2}

a + b\sqrt{2} = -21 + 5\sqrt{2}

Comparing the coefficient of

\sqrt{2}

on both sides gives

b = 5

Answer:

b = 5

(Option B).

Go to Question:

Prev Question

Next Question