Question

Solve the following quadratic equation:

`1/(x + 1) + 3/(5x + 1) = 5/(x + 4), x ≠ -1, -1/5, -4`

Answer 1

Given: `1/(x + 1) + 3/(5x + 1) = 5/(x + 4), x ≠ -1, -1/5, -4`

Step-wise calculation:

1. Multiply both sides by the common denominator (x + 1)(5x + 1)(x + 4) to clear fractions.

2. After multiplying:

(5x + 1)(x + 4) + 3(x + 1)(x + 4) = 5(x + 1)(5x + 1)

3. Expand both sides:

(5x + 1)(x + 4) = 5x² + 21x + 4

3(x + 1)(x + 4) = 3x² + 15x + 12

Sum (left) = 8x² + 36x + 16

Right: 5(x + 1)(5x + 1) = 25x² + 30x + 5

4. Bring all terms to one side:

0 = 25x² + 30x + 5 – (8x² + 36x + 16)

0 = 17x² – 6x – 11

5. Solve the quadratic 17x² – 6x – 11 = 0:

Discriminant D = (–6)² – 4 × 17 × (–11) 

= 36 + 748

= 784

`sqrt(D) = 28` 

`x = (6 ± 28)/(2 xx 17)` 

= `(6 ± 28)/34` 

So, `x = (6 + 28)/34` 

= `34/34`

= 1

or

`x = (6 - 28)/34` 

= `-22/34`

= `-11/17`

6. Check exclusions: x = 1 and x = `-11/17` are not equal to `-1, -1/5` or –4 and produce nonzero denominators, so both are valid.

The solutions are x = 1 and x = `-11/17`.