Question

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Answer 1

`2a - 3b + 4c = 1`

Explanation:

Given

`a^2 + b^2 + c^2 = 2(a - b - c) - 3`

Required

Determine
`2a - 3b + 4c`

`a^2 + b^2 + c^2 = 2(a - b - c) - 3`

Open bracket

`a^2 + b^2 + c^2 = 2a - 2b - 2c - 3`

Equate the equation to 0

`a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0`

Express 3 as 1 + 1 + 1

`a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0`

Collect like terms

`a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0`

Group each terms

`(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0`

Factorize (starting with the first bracket)

`(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0`

`(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0`

`((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0`

`((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0`

`((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0`

`((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0`

`((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0`

Express 0 as 0 + 0 + 0

`(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0`

By comparison

`(a - 1)^2 = 0`

`(b + 1)^2 = 0`

`(c + 1)^2 = 0`

Solving for
`(a - 1)^2 = 0`

Take square root of both sides

`a - 1 = 0`

Add 1 to both sides

`a - 1 + 1 = 0 + 1`

`a = 1`

Solving for
`(b + 1)^2 = 0`

Take square root of both sides

`b + 1 = 0`

Subtract 1 from both sides

`b + 1 - 1 = 0 - 1`

`b = -1`

Solving for
`(c + 1)^2 = 0`

Take square root of both sides

`c + 1 = 0`

Subtract 1 from both sides

`c + 1 - 1 = 0 - 1`

`c = -1`

Substitute the values of a, b and c in
`2a - 3b + 4c`

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

`2a - 3b + 4c = 2 +3 -4`

`2a - 3b + 4c = 1`