Math Problem Statement

12

Solution

The given equation is:

282x+5xx+3+7x2x12=0\frac{2}{8 - 2x} + \frac{5x}{x + 3} + \frac{7}{x^2 - x - 12} = 0

Let's solve this equation step by step.

Step 1: Factor the quadratic expression

The quadratic expression in the third fraction is x2x12x^2 - x - 12. We can factor this expression:

x2x12=(x4)(x+3)x^2 - x - 12 = (x - 4)(x + 3)

So the equation becomes:

282x+5xx+3+7(x4)(x+3)=0\frac{2}{8 - 2x} + \frac{5x}{x + 3} + \frac{7}{(x - 4)(x + 3)} = 0

Step 2: Simplify the first fraction

The first fraction can be simplified by factoring out the 2 from the denominator:

282x=22(4x)=14x\frac{2}{8 - 2x} = \frac{2}{2(4 - x)} = \frac{1}{4 - x}

The equation now is:

14x+5xx+3+7(x4)(x+3)=0\frac{1}{4 - x} + \frac{5x}{x + 3} + \frac{7}{(x - 4)(x + 3)} = 0

Step 3: Find a common denominator

The common denominator for all three fractions would be (4x)(x+3)(x4)(4 - x)(x + 3)(x - 4). Notice that 4x=(x4)4 - x = -(x - 4), so the common denominator can be expressed as:

(x4)(x+3)(x - 4)(x + 3)

Thus, rewrite each term with this common denominator:

(x+3)(4x)(x+3)+5x(4x)(4x)(x+3)+7(x4)(x+3)=0\frac{(x + 3)}{(4 - x)(x + 3)} + \frac{5x(4 - x)}{(4 - x)(x + 3)} + \frac{7}{(x - 4)(x + 3)} = 0

Simplifying, we get:

(x+3)5x(4x)+7(x4)(x+3)=0\frac{(x + 3) - 5x(4 - x) + 7}{(x - 4)(x + 3)} = 0

Step 4: Simplify the numerator

Let's expand and simplify the numerator:

(x+3)5x(4x)+7(x + 3) - 5x(4 - x) + 7

Expand 5x(4x)-5x(4 - x):

(x+3)(20x5x2)+7(x + 3) - (20x - 5x^2) + 7

Combine like terms:

5x2+21x+10-5x^2 + 21x + 10

So the equation now becomes:

5x2+21x+10(x4)(x+3)=0\frac{-5x^2 + 21x + 10}{(x - 4)(x + 3)} = 0

Step 5: Set the numerator equal to zero

For the equation to be zero, the numerator must be zero:

5x2+21x+10=0-5x^2 + 21x + 10 = 0

This is a quadratic equation. We can solve it using the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=5a = -5, b=21b = 21, and c=10c = 10. Plugging in these values:

x=21±2124(5)(10)2(5)x = \frac{-21 \pm \sqrt{21^2 - 4(-5)(10)}}{2(-5)} x=21±441+20010x = \frac{-21 \pm \sqrt{441 + 200}}{-10} x=21±64110x = \frac{-21 \pm \sqrt{641}}{-10}

So the roots are:

x=21+64110orx=2164110x = \frac{-21 + \sqrt{641}}{-10} \quad \text{or} \quad x = \frac{-21 - \sqrt{641}}{-10}

Simplify the expressions:

x=2164110orx=21+64110x = \frac{21 - \sqrt{641}}{10} \quad \text{or} \quad x = \frac{21 + \sqrt{641}}{10}

Step 6: Exclude extraneous solutions

We need to exclude any solutions that make the original denominators zero. The denominators were:

4x,x+3,x44 - x, \quad x + 3, \quad x - 4

These give potential exclusions x=4x = 4 and x=3x = -3. Check these values against the roots:

  • x=2164110x = \frac{21 - \sqrt{641}}{10} is not 4 or -3.
  • x=21+64110x = \frac{21 + \sqrt{641}}{10} is not 4 or -3.

So both roots are valid.

Final Answer:

The solutions are:

x=2164110,x=21+64110x = \frac{21 - \sqrt{641}}{10}, \quad x = \frac{21 + \sqrt{641}}{10}

Would you like more details or have any questions?


Here are some additional questions to consider:

  1. What is the importance of factoring quadratic expressions in solving equations?
  2. How does the quadratic formula help in solving any quadratic equation?
  3. Why do we need to check for extraneous solutions?
  4. What are the possible implications of having a negative discriminant in the quadratic formula?
  5. How can you simplify complex fractions involving polynomials?

Tip: When solving quadratic equations, always simplify the expression as much as possible before applying the quadratic formula to avoid unnecessary complexity.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Factoring

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

High School