WAEC 2017 · Paper 2 · Q12

(a)
Using the method of completing the square, solve, correct to 2 decimal places, the equation $3y^2 - 5y + 2 = 0$ .
Separate values with commas, e.g. 3, −2
(b)
Given that $M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}$ , $N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}$ and $MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ , find the matrix $N$ .
Show the answer
$N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Worked solution (try it first)

Divide by 3 so that $y^2$ stands alone, and move the number across: $y^2 - \frac53y = -\frac23$ .
Half of $-\frac53$ is $-\frac56$ .
Add its square, $\frac{25}{36}$ , to both sides: $\left(y - \frac56\right)^2 = \frac{25}{36} - \frac{24}{36}$
$= \frac{1}{36}$ .
Take square roots: $y - \frac56 = \pm\frac16$ , so $y = \frac56 + \frac16 = 1$ or $y = \frac56 - \frac16 = \frac23$ .
To 2 decimal places, $y = 1.00$ or $y = 0.67$ .

Multiply row by column: $MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix}$ .
Match each entry with $\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ .
The first column gives $m + 2n = 2$ and $4m + 3n = 3$ .
Take 4 times the first from the second: $-5n = -5$ , so $n = 1$ and then $m = 0$ .
The second column gives $x + 2y = 1$ and $4x + 3y = 4$ .
In the same way, $-5y = 0$ , so $y = 0$ and then $x = 1$ .
So $N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ .
Check: $\begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ ✓.