Question

If `a = 1/(4 - a)`, find the value of `a^2 + 1/a^2`.

Answer 1

Given: `a = 1/(4 - a)`

Step-wise calculation:

1. Multiply both sides by (4 – a) to get rid of the denominator:

a(4 – a) = 1

4a – a² = 1

2. Rearrange to form a quadratic equation:

a² – 4a + 1 = 0

3. Solve the quadratic using the formula:

`a = (4 ± sqrt((-4)^2 - 4 xx 1 xx 1))/2`

`a = (4 ± sqrt(16 - 4))/2`

`a = (4 ± sqrt(12))/2`

`a = (4 ± 2sqrt(3))/2`

`a = 2 ± sqrt(3)`

4. Calculate `a + 1/a` first:

Since `a = 2 ± sqrt(3)`,

`1/a = 1/(2 ± sqrt(3))`

Rationalize denominator:

`1/(2 ± sqrt(3)) xx (2 -+ sqrt(3))/(2 -+ sqrt(3)) = (2 -+ sqrt(3))/((2)^2 - (sqrt(3))^2`

`1/(2 ± sqrt(3)) xx (2 -+ sqrt(3))/(2 -+ sqrt(3)) = (2 -+ sqrt(3))/(4 - 3)`

`1/(2 ± sqrt(3)) xx (2 -+ sqrt(3))/(2 -+ sqrt(3)) = 2 -+ sqrt(3)`

5. Hence,

`a + 1/a = (2 ± sqrt(3)) + (2 -+ sqrt(3))`

`a + 1/a = 2 + 2`

`a + 1/a = 4`

6. Use the identity:

`(a + 1/a)^2 = a^2 + 2 + 1/a^2`

So, `a^2 + 1/a^2 = (a + 1/a)^2 - 2`

`a^2 + 1/a^2 = 4^2 - 2`

`a^2 + 1/a^2 = 16 - 2`

`a^2 + 1/a^2 = 14`