Solve each of the following quadratic equations:
(i) `(1)/(x+1)+(2)/(x+2)=(5)/(x+4),xne-1,-2,-4`
(ii) `(1)/(x+1)+(3)/(5x+1)=(5)/(x+4),xne-1,-(1)/(5),-4`

Answer

To solve the given quadratic equations, we will follow a systematic approach. Let's break down the solutions step by step.

### (i) Solve the equation:
\[ \frac{1}{x+1} + \frac{2}{x+2} = \frac{5}{x+4} \]

**Step 1: Find the LCM of the denominators.**
The denominators are \(x+1\), \(x+2\), and \(x+4\). The LCM is:
\[ (x+1)(x+2)(x+4) \]

**Step 2: Multiply through by the LCM to eliminate the fractions.**
\[ (x+1)(x+2)(x+4) \left(\frac{1}{x+1} + \frac{2}{x+2}\right) = (x+1)(x+2)(x+4) \cdot \frac{5}{x+4} \]
This simplifies to:
\[ (x+2)(x+4) + 2(x+1)(x+4) = 5(x+1)(x+2) \]

**Step 3: Expand both sides.**
Left side:
\[ (x+2)(x+4) + 2(x+1)(x+4) = x^2 + 6x + 8 + 2(x^2 + 5x + 4) = x^2 + 6x + 8 + 2x^2 + 10x + 8 = 3x^2 + 16x + 16 \]
Right side:
\[ 5(x+1)(x+2) = 5(x^2 + 3x + 2) = 5x^2 + 15x + 10 \]

**Step 4: Set the equation to zero.**
\[ 3x^2 + 16x + 16 - (5x^2 + 15x + 10) = 0 \]
This simplifies to:
\[ -2x^2 + x + 6 = 0 \]
Multiplying through by -1 gives:
\[ 2x^2 - x - 6 = 0 \]

**Step 5: Apply the quadratic formula.**
The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = -1\), and \(c = -6\).
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2} \]
\[ x = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4} \]

**Step 6: Calculate the roots.**
1. \(x = \frac{8}{4} = 2\)
2. \(x = \frac{-6}{4} = -\frac{3}{2}\)

### Solution for (i):
The solutions are:
\[ x = 2 \quad \text{and} \quad x = -\frac{3}{2} \]

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### (ii) Solve the equation:
\[ \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4} \]

**Step 1: Find the LCM of the denominators.**
The denominators are \(x+1\), \(5x+1\), and \(x+4\). The LCM is:
\[ (x+1)(5x+1)(x+4) \]

**Step 2: Multiply through by the LCM to eliminate the fractions.**
\[ (x+1)(5x+1)(x+4) \left(\frac{1}{x+1} + \frac{3}{5x+1}\right) = (x+1)(5x+1)(x+4) \cdot \frac{5}{x+4} \]
This simplifies to:
\[ (5x+1)(x+4) + 3(x+1)(x+4) = 5(x+1)(5x+1) \]

**Step 3: Expand both sides.**
Left side:
\[ (5x+1)(x+4) + 3(x+1)(x+4) = (5x^2 + 20x + x + 4) + 3(x^2 + 5x + 4) = 5x^2 + 21x + 4 + 3x^2 + 15x + 12 = 8x^2 + 36x + 16 \]
Right side:
\[ 5(x+1)(5x+1) = 5(5x^2 + 6x + 1) = 25x^2 + 30x + 5 \]

**Step 4: Set the equation to zero.**
\[ 8x^2 + 36x + 16 - (25x^2 + 30x + 5) = 0 \]
This simplifies to:
\[ -17x^2 + 6x + 11 = 0 \]
Multiplying through by -1 gives:
\[ 17x^2 - 6x - 11 = 0 \]

**Step 5: Apply the quadratic formula.**
Here, \(a = 17\), \(b = -6\), and \(c = -11\).
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 17 \cdot (-11)}}{2 \cdot 17} \]
\[ x = \frac{6 \pm \sqrt{36 + 748}}{34} = \frac{6 \pm \sqrt{784}}{34} = \frac{6 \pm 28}{34} \]

**Step 6: Calculate the roots.**
1. \(x = \frac{34}{34} = 1\)
2. \(x = \frac{-22}{34} = -\frac{11}{17}\)

### Solution for (ii):
The solutions are:
\[ x = 1 \quad \text{and} \quad x = -\frac{11}{17} \]

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