Question

Solve the equation if a²+b²+c²=2(a-b-c)-3 then find out the value of 2a-3b+4c=?

Answer 1

Answer:

1

Step-by-step explanation:

a² + b² + c² = 2 (a - b - c ) - 3

=> a² + b² + c² = 2a - 2b - 2c -3

=> a² + b² + c² - 2a + 2b + 2c + 3 = 0

//rearrange the above as below

=> a² - 2a + 1 + b² + 2b + 1  + c² + 2c + 1 = 0

=> (a-1)² + (b+1)² + (c+1)² = 0

Using sum of all three real quantity is zero, if all are zero

=>  a - 1  = 0 => a = 1

     b + 1  = 0  => b = -1

     c + 1 = 0 => c = - 1

Thus value of 2a - 3b + 4c = 2 * 1 - 3(-1) + 4(-1) = 1

Answer 2

Answer:

1

Step-by-step explanation:

$a^{2} + b^{2} + c^{2} = 2(a-b-c) - 3\\a^{2} + b^{2} + c^{2} = 2a - 2b - 2c - 3\\a^{2} + b^{2} + c^{2} - 2a + 2b + 2c + 3 = 0$

Rearrange in this way so that squares can be formed,

$(a^{2} - 2a + 1) + (b^{2} + 2b + 1) + (c^{2} + 2c + 1) = 0\\(a - 1)^{2} + (b + 1)^{2} + (c + 1)^{2} = 0$

Since the squares are equal to 0, each term is also equal to zero.

Hence,

a - 1 = 0

b + 1 = 0

c + 1 = 0

a = 1

b = -1

c = -1

Therefore, 2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)

                                       = 2 + 3 - 4

                                       = 1