Question

a^2+4b^2+49c^2+18=2(2b+28c-a), then the value of (3a+2b+7c)

Answer 1

(3a + 2b + 7c) = 2

it is given that, a² + 4b² + 49c² + 18 = 2(2b + 28c - a)

⇒a² + 4b² + 49c² + 18 = 4b + 56c - 2a

⇒a² + 2a + 4b² - 4b + 49c² - 56c + 18 = 0

⇒a² + 2a + 1 + 4b² - 4b + 1 + 49c² - 56c + 16 = 0

[ use formula, a² - 2ab + b² = (a - b)² ]

⇒(a + 1)² + (2b - 1)² + (7c - 4)² = 0

⇒a = -1, b = 1/2 , c = 4/7

then, (3a + 2b + 7c)

= 3 × -1 + 2 × 1/2 + 7 × 4/7

= -3 + 1 + 4

= 2

also read similar questions: Factorise: (a+2b)^2-(a+2b)(3a-7b)

https://brainly.in/question/6853962

Solve the following quadratic equation in x :

a^2b^2x^2 - 4b^4x + 3a^4x - 12a^2b^2 = 0

https://brainly.in/question/8403748

Answer 2

Question :------- if a²+4b²+49c²+18 = 2(2b+28c-a)

Find the value of (3a+2b+7c) ?

Formula to be used :---

• a²+b²-2ab = (a-b)²
• re- arrangement ..

a²+4b²+49c²+18 = 2(2b+28c-a)

a²+4b²+49c²+18 = 4b + 56c -2a

a²+4b²+49c²+18 - 4b - 56c + 2a = 0

Re - arranging them now ,

(a²+2a+1)+(4b²-4b+1) + (49c²-56c +16) = 0

(a+1)² + (2b-1)² + (7c-4)² = 0

a = - 1,

b = 1/2

c = 4/7

Putting values now we get,

(3a+2b+7c) = 3*(-1)+2*1/2+7*4/7 = -3 + 1 + 4 = 2 (Ans)

(Hope it helps you)