If (6a2 – 5ab – 4b2) = 0 and (144a2 + 324b2 + 432ab – 289) = 0, then what will be the value of (a2 + b2) : (a2 – b2) where both ‘a’ and ‘b’ are positive numbers?

If (6a2 – 5ab – 4b2) = 0 and (144a2 + 324b2 + 432ab – 289) = 0, then what will be the value of (a2 + b2) : (a2 – b2) where both ‘a’ and ‘b’ are positive numbers? Correct Answer 25 : 7

GIVEN:

(6a² – 5ab – 4b²) = 0 and (144a² + 324b² + 432ab – 289) = 0

CONCEPT:

Application of algebraic formulas.

FORMULA USED:

(a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

CALCULATION:

(6a² – 5ab – 4b²) = 0

⇒ (6a² + 3ab – 8ab – 4b²) = 0

⇒ 3a (2a + b) – 4b (2a + b) = 0

⇒ (2a + b)(3a – 4b) = 0

If 2a + b = 0, then either of the two number should be negative. So, we will not consider this case.

⇒ (3a – 4b) = 0     ---- (1)

(144a² + 324b² + 432ab – 289) = 0

⇒ 36(4a² + 9b² + 12ab) = 289

⇒ (2a + 3b)² = 289/36

⇒ (2a + 3b) = 17/6      ---- (2)

From (1) and (2):

a = 2/3

⇒ a² = 4/9

b = 1/2

⇒ b² = 1/4

Now,

(a² + b²) = (4/9) + (1/4) = (25/36)

(a² – b²) = (4/9) – (1/4) = (7/36)

Hence,