Question

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

• A. z > k
• B. z < k
• C. z = k or the relationship cannot be established
• D. z ≤ k

Answer 1

The Correct Option is D

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 × 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
So, the correct option is (D) : z ≤ k.