Question

1. \\(\frac{x^2 + 13x + 40}{x^2 - 2x - 8} - \frac{1}{x + 2} = \frac{4x^2 + 36x + 80}{x^2 - 2x - 8}\\)

Explanation:

Step1: Factor Denominators

Factor \(x^2 - 2x - 8=(x - 4)(x + 2)\), \(x^2+13x + 40=(x + 5)(x + 8)\), \(4x^2+36x + 80 = 4(x^2+9x + 20)=4(x + 4)(x + 5)\). The equation becomes \(\frac{(x + 5)(x + 8)}{(x - 4)(x + 2)}-\frac{1}{x + 2}=\frac{4(x + 4)(x + 5)}{(x - 4)(x + 2)}\).

Step2: Multiply by LCD

The least common denominator (LCD) is \((x - 4)(x + 2)\). Multiply each term by LCD: \((x + 5)(x + 8)- (x - 4)=4(x + 4)(x + 5)\).

Step3: Expand Expressions

Expand left side: \(x^2+13x + 40 - x + 4=x^2+12x + 44\). Expand right side: \(4(x^2+9x + 20)=4x^2+36x + 80\).

Step4: Simplify Equation

Bring all terms to one side: \(x^2+12x + 44-(4x^2+36x + 80)=0\) → \(-3x^2-24x - 36 = 0\) → Divide by -3: \(x^2+8x + 12 = 0\).

Step5: Solve Quadratic

Factor \(x^2+8x + 12=(x + 2)(x + 6)=0\). Solutions: \(x=-2\) or \(x=-6\).

Step6: Check Extraneous Solutions

\(x=-2\) makes denominator \(x + 2 = 0\) and \(x^2 - 2x - 8 = 0\), so it's extraneous. \(x=-6\) is valid.

Answer:

\(x = -6\)