Question

6 (a + b)2 x2 + 5 (a + b)cx – 6c2 = 0
solve by completing the square​

Answer 1

6(a + b)²x² + 5(a + b)cx - 6c² = 0

⇒(a + b)²x² + (5/6)(a + b)cx - c² = 0

⇒{(a + b)x}² + 2.(5c/12).(a + b) x - c² = 0

adding (5c/12)² both sides,

⇒{(a + b)x}² + 2(5c/12).(a + b)x + (5c/12)² - c² = (5c/12)²

⇒{(a + b)x + 5c/12}² - c² = (5c/12)²

⇒{(a + b)x + 5c/12}² = c² + (5c/12)²

⇒{(a + b)x + 5c/12}² = (144 + 25)c²/144

⇒{(a + b)x + 5c/12}² = (13c/12)²

taking square root both sides,

⇒(a + b)x + 5c/12 = ± 13c/12

⇒(a + b)x = (-5c ± 13c)/12

⇒x = (-5c ± 13c)/12(a + b)

hence, x = -3c/2(a + b) , 2/3(a + b)

Answer 2

Answer:

Refer to the attachment

Step-by-step explanation:

hope its help u