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 f... 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 = x^2 - 11x - 5 II. 5y(y - 3) - 64 = y(3y - 2) - 19 III. (z + 2a - d)^2 = 169 Read more

Steps to Solve

1. Solve Equation I for x
   The first step is to rearrange and solve equation I:
   $$ 3x^2 - 36 + 72 - 11x - 5 = 0 $$ This simplifies to:
   $$ 3x^2 - 11x + 31 = 0 $$

2. Use the Quadratic Formula
   We apply the quadratic formula:
   $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
   Where ( a = 3, b = -11, c = 31 ):
   Calculate the discriminant:
   $$ b^2 - 4ac = (-11)^2 - 4(3)(31) = 121 - 372 = -251 $$ Since the discriminant is negative, there are no real roots.

3. Solve Equation II for y
   Rearranging equation II gives:
   $$ 5y^2 - 15y - 64 + 19 = 0 $$
   This simplifies to:
   $$ 5y^2 - 15y - 45 = 0 $$
   Dividing through by 5:
   $$ y^2 - 3y - 9 = 0 $$

4. Use the Quadratic Formula for y
   By applying the quadratic formula again:
   $$ y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-9)}}{2(1)} $$
   Calculate the discriminant:
   $$ 9 + 36 = 45 $$
   Roots are:
   $$ y = \frac{3 \pm \sqrt{45}}{2} $$
   Therefore, roots are real.

5. Analyze Equation III
   Given:
   $$ (z + 2a - d)^2 = 169 $$
   Taking the square root gives two scenarios:
   $$ z + 2a - d = 13 \quad \text{or} \quad z + 2a - d = -13 $$
   Rearranging gives:
   $$ z = 13 - 2a + d \quad \text{or} \quad z = -13 - 2a + d $$

6. Establish the Relationship
   Since ( k = 11 ), compare:
   $$ 13 - 2a + d \quad \text{and} \quad -13 - 2a + d $$
   This establishes that ( z ) can vary depending on ( a ) and ( d ) and leads to ( z \geq k ) or ( z \leq k ).