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}).