(a) Using completing the square method, solve, correct to 2 decimal places, the equation \(3y^{2}

Explanation

(a) \(3y^{2} - 5y + 2 = 0\)

\(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\)

\(y^{2} - \frac{5}{3}y = \frac{-2}{3}\)

\(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\)

\(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\)

\((y - (\frac{5}{6}))^{2} = \frac{1}{36}\)

Taking square root of both sides,

\(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\)

\(y - \frac{5}{6} =\pm {\frac{1}{6}}\)

\(y = \frac{5}{6} \pm \frac{1}{6}\)

\(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\).

(b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\)

\(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\)

\(m + 2n = 2 ... (1)\)

\(4m + 2n = 3 ... (2)\)

Multiplying (1) by 4, we have

\(4m + 8n = 8 ... (3)\)

(3) - (1) : \(8n - 2n = 8 - 2 = 6\)

\(6n = 6 \implies n = 1\)

\(m + 2n = m + 2(1) = m + 2 = 2\)

\(m = 2 - 2 = 0\)

\(x + 2y = 1 ... (1)\)

\(4x + y = 4 ... (2)\)

Multiply (1) by 4,

\(4x + 8y = 4 ... (3)\)

(3) - (2) : \(8y - y = 4 - 4 = 0\)

\(\implies y = 0\)

\(x + 2y = x + 2(0) = x + 0 = 1\)

\(x = 1 - 0 = 1\)

\(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).