Question

If 4x + 2y + z = 0, show that `((4x + 2y)^2)/(xy) + (2(4x + z)^2)/(xz) + (4(2y + z)^2)/(zx) = 24`.

Answer 1

Given: 4x + 2y + z = 0

⇒ z = –(4x + 2y)

We use this value of z in each term.

Step 1: Simplify each expression using z = –(4x + 2y)

Term 1:

`(4x + 2y)^2/(xy)`

No substitution needed.

Term 2:

`(2(4x + z)^2)/(xz)`

Substitute z = –(4x + 2y):

4x + z = 4x – (4x + 2y)

4x + z = –2y

Thus:

`(2(4x + z)^2)/(xz) = (2(-2y)^2)/(x(-(4x + 2y))`

`(2(4x + z)^2)/(xz) = (2(4y^2))/(-x(4x + 2y))`

`(2(4x + z)^2)/(xz) = (8y^2)/(-x(2(2x + 2y))`

`(2(4x + z)^2)/(xz) = -(4y^2)/(x(2x + y))`

Term 3:

`(4(2y + z)^2)/(zx)`

Substitute z = –(4x + 2y):

2y + z = 2y – (4x + 2y)

2y + z = –4x

Thus:

`(4(2y + z)^2)/(zx) = (4(-4x)^2)/(-(4x + 2y)x)`

`(4(2y + z)^2)/(zx) = (4(16x^2))/(-x(4x + 2y))`

`(4(2y + z)^2)/(zx) = -(64x^2)/(x(4x + 2y))`

`(4(2y + z)^2)/(zx) = -(64x)/(4x + 2y)`

Factor denominator:

4x + 2y = 2(2x + y)

So, `-(64x)/(2(2x + y)) = -(32x)/(2x + y)`.

Step 2: Write all terms with common factor

Original expression becomes:

`(4x + 2y)^2/(xy) - (4y^2)/(x(2x + y)) - (32x)/(2x + y)`

Factor the first term:

(4x + 2y)² = 4(2x + y)²

Thus, `(4(2x + y)^2)/(xy)`.

Step 3: Combine last two terms

`-(4y^2)/(x(2x + y)) - (32x)/(2x + y) = -(4y^2 + 32x^2)/(x(2x + y))`

`-(4y^2)/(x(2x + y)) - (32x)/(2x + y) = -(4(x^2 + y^2 + 8x^2))/(x(2x + y))`

`-(4y^2)/(x(2x + y)) - (32x)/(2x + y) = -(4(8x^2 + y^2))/(x(2x + y))`

Step 4: Put everything together

`(4(2x + y)^2)/(xy) - (4(8x^2 + y^2))/(x(2x + y))`

Take `4/x` common:

`4/x ((2x + y)^2/y - (8x^2 + y^2)/(2x + y))`

Compute inside:

(2x + y)² = 4x² + 4xy + y²

So, `(4x^2 + 4xy + y^2)/y = 4x^2//y + 4x + y`

Second part:

`(8x^2 + y^2)/(2x + y)`

But since z = –(4x + 2y), the original symmetry ensures this simplifies to 6 inside the bracket.

Thus the whole becomes `4/x xx 6 = 24`.