WAEC 2014 · Paper 2 · Q10

(a)
Express $\dfrac{5 + \sqrt2}{3 - \sqrt2} - \dfrac{5 - \sqrt2}{3 + \sqrt2}$ in the form $a + b\sqrt2$ .
(b)
Solve the following equations simultaneously using the determinant method:

$3x - y - z = -2$

$x + 5y + 2z = 5$

$2x + 3y + z = 0$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Use the common denominator $(3 - \sqrt2)(3 + \sqrt2) = 9 - 2 = 7$ .
First numerator: $(5 + \sqrt2)(3 + \sqrt2) = 15 + 5\sqrt2 + 3\sqrt2 + 2$
$= 17 + 8\sqrt2$ .
Second numerator: $(5 - \sqrt2)(3 - \sqrt2) = 15 - 5\sqrt2 - 3\sqrt2 + 2$
$= 17 - 8\sqrt2$ .
Subtract: $\dfrac{(17 + 8\sqrt2) - (17 - 8\sqrt2)}{7} = \dfrac{16\sqrt2}{7}$ .
In the form $a + b\sqrt2$ : $0 + \frac{16}{7}\sqrt2$ , so $a = 0$ and $b = \frac{16}{7}$ .

$\Delta = \begin{vmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{vmatrix}$
$= 3(5 - 6) + 1(1 - 4) - 1(3 - 10)$ .
So $\Delta = -3 - 3 + 7 = 1$ .
Replace the first column by $(-2, 5, 0)$ : $\Delta_x = -2(5 - 6) + 1(5 - 0) - 1(15 - 0)$
$= -8$ .
Replace the second column: $\Delta_y = \begin{vmatrix} 3 & -2 & -1 \\ 1 & 5 & 2 \\ 2 & 0 & 1 \end{vmatrix}$
$= 15 - 6 + 10$
$= 19$ .
Replace the third column: $\Delta_z = \begin{vmatrix} 3 & -1 & -2 \\ 1 & 5 & 5 \\ 2 & 3 & 0 \end{vmatrix}$
$= -45 - 10 + 14$
$= -41$ .
Divide each by $\Delta = 1$ : $x = -8$ , $y = 19$ and $z = -41$ .
Check in the third equation: $2(-8) + 3(19) + (-41) = 0$ ✓.