How to Solve & Graph a Rational Inequality with Quadratics in the Numerator & Denominator | Tutorial

Introduction to Rational Inequality

  • Focus of the video tutorial: solving and graphing a given rational inequality.

  • First crucial step: ensuring 0 is on one side of the inequality, which is verified in this case.

Denominator Analysis

  • Denominator: ( x^2 - 9 )

    • Recognized as the difference of squares: ( x^2 - 3^2 )

    • Factored: ( (x - 3)(x + 3) )

Numerator Analysis

  • Numerator: ( x^2 + x + 6 )

    • Discriminant calculation: ( b^2 - 4ac )

      • Coefficients: ( a = 1, b = 1, c = 6 )

      • Calculation: ( 1^2 - 4 \times 1 \times 6 = 1 - 24 = -23 )

    • Interpretation: The quadratic is not factorable due to a negative discriminant.

Inequality Structure

  • Final form of the inequality: ( \frac{x^2 + x + 6}{(x - 3)(x + 3)} > 0 )

Finding Critical Values

  • Setting numerator and denominator to zero:

    • Numerator: ( x^2 + x + 6 = 0 ) yields no real solutions (discriminant < 0).

    • Denominator:

      • ( x - 3 = 0 ) gives ( x = 3 )

      • ( x + 3 = 0 ) gives ( x = -3 )

  • Critical points identified: ( x = -3 ) and ( x = 3 )

Number Line Representation

  • Placement of critical points on a number line:

    • ( x = -3 ) represented with a dotted line (open circle).

    • ( x = 3 ) represented with a dotted line (open circle).

Dividing into Regions

  • Regions defined by critical points:

    • Region A: ( (-\infty, -3) )

    • Region B: ( (-3, 3) )

    • Region C: ( (3, +\infty) )

Testing Points in Regions

  • Region A (choose ( x = -4 )):

    • Calculation: ( \frac{(-4)^2 + (-4) + 6}{(-4 - 3)(-4 + 3)} )

    • Result: Positive (true for Region A).

  • Region B (choose ( x = 0 )):

    • Calculation: ( \frac{0^2 + 0 + 6}{(0 - 3)(0 + 3)} )

    • Result: Negative (false for Region B).

  • Region C (choose ( x = 4 )):

    • Calculation: ( \frac{4^2 + 4 + 6}{(4 - 3)(4 + 3)} )

    • Result: Positive (true for Region C).

Shading Valid Regions

  • Regions to shade based on test results:

    • Shade Region A (true).

    • Ignore Region B (false).

    • Shade Region C (true).

Final Representation

  • Resulting interval notation for the solution:

    • ( (-\infty, -3) \cup (3, +\infty) ) (both ends excluded, hence parentheses).

  • Graph representation includes shaded areas indicating true regions.

Conclusion

  • Important note on union in notation: can be interpreted as “or.”

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