Math Problem Statement

Factorize

Solution

The new equation from the image is:

9x17x1x1x+3=12\frac{9}{x-1} - \frac{7}{x-1} - \frac{x-1}{x+3} = \frac{1}{2}

Let's solve this equation step-by-step:

  1. Simplify the equation:

    Combine the first two fractions on the left side that have the same denominator:

97x1x1x+3=12\frac{9 - 7}{x-1} - \frac{x-1}{x+3} = \frac{1}{2}

This simplifies to:

2x1x1x+3=12\frac{2}{x-1} - \frac{x-1}{x+3} = \frac{1}{2}

  1. Combine the fractions on the left side:

    The common denominator for the fractions on the left side is (x1)(x+3)(x-1)(x+3).

2(x+3)(x1)2(x1)(x+3)=12\frac{2(x+3) - (x-1)^2}{(x-1)(x+3)} = \frac{1}{2}

  1. Expand and simplify the numerator:

    • Distribute the 2 in the first term:

2x+6(x1)2(x1)(x+3)=12\frac{2x + 6 - (x-1)^2}{(x-1)(x+3)} = \frac{1}{2}

  • Expand (x1)2(x-1)^2:

(x1)2=x22x+1(x-1)^2 = x^2 - 2x + 1

So, the equation becomes:

2x+6(x22x+1)(x1)(x+3)=12\frac{2x + 6 - (x^2 - 2x + 1)}{(x-1)(x+3)} = \frac{1}{2}

  1. Simplify further:

Combine like terms in the numerator:

2x+6x2+2x1(x1)(x+3)=12\frac{2x + 6 - x^2 + 2x - 1}{(x-1)(x+3)} = \frac{1}{2}

This simplifies to:

x2+4x+5(x1)(x+3)=12\frac{-x^2 + 4x + 5}{(x-1)(x+3)} = \frac{1}{2}

  1. Clear the fraction by multiplying both sides by 2(x1)(x+3)2(x-1)(x+3):

Multiply both sides of the equation by 2(x1)(x+3)2(x-1)(x+3) to eliminate the fraction:

2(x2+4x+5)=(x1)(x+3)2(-x^2 + 4x + 5) = (x-1)(x+3)

  1. Expand the right-hand side:

Expand (x1)(x+3)(x-1)(x+3):

(x1)(x+3)=x2+2x3(x-1)(x+3) = x^2 + 2x - 3

So the equation becomes:

2x2+8x+10=x2+2x3-2x^2 + 8x + 10 = x^2 + 2x - 3

  1. Move all terms to one side of the equation:

2x2x2+8x2x+10+3=0-2x^2 - x^2 + 8x - 2x + 10 + 3 = 0

Simplify:

3x2+6x+13=0-3x^2 + 6x + 13 = 0

  1. Factorize or use the quadratic formula:

This quadratic equation can be solved using the quadratic formula:

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

Where a=3a = -3, b=6b = 6, and c=13c = 13.

Plug in the values:

x=6±624(3)(13)2(3)x = \frac{-6 \pm \sqrt{6^2 - 4(-3)(13)}}{2(-3)}

x=6±36+1566x = \frac{-6 \pm \sqrt{36 + 156}}{-6}

x=6±1926x = \frac{-6 \pm \sqrt{192}}{-6}

Simplify:

x=6±836x = \frac{-6 \pm 8\sqrt{3}}{-6}

This gives the solutions:

x=6±836x = \frac{6 \pm 8\sqrt{3}}{6}

Or:

x=1±433x = 1 \pm \frac{4\sqrt{3}}{3}

These are the roots of the quadratic equation.

Would you like any further details or clarifications on these steps?

Here are some related questions:

  1. How can you verify the solutions to a quadratic equation?
  2. Why is it important to simplify the equation before solving?
  3. What other methods can be used to solve quadratic equations?
  4. How do you determine the best method to solve a quadratic equation?
  5. What are some practical applications of solving quadratic equations?

Tip: Always simplify the equation as much as possible before solving, which often makes the solution process easier.

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Math Problem Analysis

Mathematical Concepts

Equations
Fractions
Quadratic Equations

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

High School