If 9a^2 + 16b^2 + c^2 + 25 = 24 (a + b) then find 3a+4b+5c.
Answers
answer : 7
explanation : it is given that,
9a² + 16b² + c² + 25 = 24(a + b)
⇒9a² + 16b² + c² + 25 - 24a - 24b = 0
⇒(3a)² - 24a + 16 + (4b)² - 24b + 9 + c² + = 0
⇒(3a)² - 2(3a)(4) + (4)² + (4b)² - 2(4b)(3) + 3² + c² = 0
using formula,
x² - 2xy + y² = (x - y)²
(3a)² - 2(3a)(4) + (4)² = (3a - 4)²
(4b)² - 2(4b)(3) + (4)² = (4b - 3)²
now, (3a)² - 2(3a)(4) + (4)² + (4b)² - 2(4b)(3) + 3² + c² = (3a - 4)² + (4b - 3)² + c² = 0
⇒(3a - 4)² + (4b - 3)² + c² = 0
this is possible only when
3a = 4, 4b = 3 and c = 0
so, 3a + 4b + 5c = 4 + 3 + 0 = 7
hence, value of 3a + 4b + 5c = 7
it is given that,
9a² + 16b² + c² + 25 = 24(a + b)
⇒9a² + 16b² + c² + 25 - 24a - 24b = 0
⇒(3a)² - 24a + 16 + (4b)² - 24b + 9 + c² + = 0
⇒(3a)² - 2(3a)(4) + (4)² + (4b)² - 2(4b)(3) + 3² + c² = 0
By using formula,
x² - 2xy + y² = (x - y)²
(3a)² - 2(3a)(4) + (4)² = (3a - 4)²
(4b)² - 2(4b)(3) + (4)² = (4b - 3)²
now, (3a)² - 2(3a)(4) + (4)² + (4b)² - 2(4b)(3) + 3² + c² = (3a - 4)² + (4b - 3)² + c² = 0
⇒(3a - 4)² + (4b - 3)² + c² = 0
this is possible only when
3a = 4, 4b = 3 and c = 0
so, 3a + 4b + 5c = 4 + 3 + 0 = 7