Given below are 3 equations I, II and III where 'a' and 'b' are the roots of equation I where (a < b) and 'c' and’d’ are roots of equation II where (c < d). On this basis, solve for equation III and find the relationship between 'z' and 'k' given that k = 11 I. 3x(x - 12) + 72 = x²- 11x – 5 II. 5y(y - 3) - 64 = y(3y - 2) – 19 III. (z + 2a - d)² = 169

Explanation

5 Detailed solution: From I: 3x(x - 12) + 72 = x²- 11x – 5 Or, 3x²- 36x + 72 – x²+ 11x + 5 = 0 Or, 2x²- 25x + 77 = 0 Or, 2x²- 14x - 11x + 77 = 0 Or, 2x(x - 7)-11(x - 7) = 0 Or, (2x - 11) (x - 7) = 0 So, x = 5.5 or 7 So, a = 5.5 and b = 7 From II: 5y(y - 3) - 64 = y(3y - 2) – 19 Or, 5y²- 15y - 64 = 3y²- 2y – 19 Or, 5y²- 3y²- 15y + 2y - 64 + 19 = 0 Or, 2y²- 13y - 45 = 0 Or, 2y²- 18y + 5y - 45 = 0 Or, 2y(y - 9) + 5(y - 9) = 0 Or, (2y + 5) (y - 9) = 0 So, y = -(5/2) = -2.5 or 9 So, c = -2.5 and d = 9 From III: (z + 2a - d)² = 169 Or, (z + 2 X 5.5 - 9)² = 169 Or, (z + 2)² = 169 Or, z + 2 = ±13 So, z = 11 or -15 ‘z’ Relation ‘k’ 11 = 11 -15 < 11 Therefore, z ≤ k