Question

If 9a^2+ 16b^2 +c^2 +25 =24{a+b} then value of 3a+4b+5c in numeric value is?

Answer 1

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

Answer 2

$\bf{\underline{\underline{Answer:}}}$

∴ $\bf{value\: of\: 3a + 4b + 5c = 7}$

$\bf{\underline{\underline{Step\: by\: step\: explanation:}}}$

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

∴ value of 3a + 4b + 5c = 7