Quadratic Formula Cut And Paste Answers 6-10

Understanding and Applying the Quadratic Formula: A Step-by-Step Guide to Solving Problems 6–10

The quadratic formula is a cornerstone of algebra, offering a systematic way to solve quadratic equations of the form $ ax^2 + bx + c = 0 $. Because of that, while factoring or completing the square can sometimes work, the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ guarantees a solution for any quadratic equation. In this article, we’ll explore problems 6–10, which test your ability to apply this formula accurately. Whether you’re a student tackling homework or a lifelong learner brushing up on math skills, mastering these examples will solidify your understanding of quadratic equations.

Problem 6: Solving $ x^2 - 5x + 6 = 0 $

Step 1: Identify coefficients
For the equation $ x^2 - 5x + 6 = 0 $, the coefficients are:

• $ a = 1 $ (coefficient of $ x^2 $)
• $ b = -5 $ (coefficient of $ x $)
• $ c = 6 $ (constant term)

Step 2: Plug into the quadratic formula
Substitute $ a $, $ b $, and $ c $ into $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:
$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} $

Step 3: Simplify the discriminant
The discriminant $ b^2 - 4ac = 25 - 24 = 1 $. Since it’s positive, there are two real solutions:
$ x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2} $

Step 4: Calculate the roots

• $ x = \frac{5 + 1}{2} = 3 $
• $ x = \frac{5 - 1}{2} = 2 $

Answer: The solutions are $ x = 3 $ and $ x = 2 $.

Problem 7: Solving $ 2x^2 + 3x - 2 = 0 $

Step 1: Identify coefficients

• $ a = 2 $
• $ b = 3 $
• $ c = -2 $

Step 2: Apply the quadratic formula
$ x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} $

Step 3: Simplify the discriminant
$ 9 + 16 = 25 $, so:
$ x = \frac{-3 \pm 5}{4} $

Step 4: Solve for $ x $

• $ x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2} $
• $ x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2 $

Answer: The solutions are $ x = \frac{1}{2} $ and $ x = -2 $.

Problem 8: Solving $ x^2 + 4x + 4 = 0 $

Step 1: Identify coefficients

• $ a = 1 $
• $ b = 4 $
• $ c = 4 $

Step 2: Apply the quadratic formula $ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 16}}{2} $

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Step 3: Simplify the discriminant $ 16 - 16 = 0 $, so: $ x = \frac{-4 \pm \sqrt{0}}{2} = \frac{-4 \pm 0}{2} $

Step 4: Solve for x

• $ x = \frac{-4}{2} = -2 $ Since the discriminant is zero, there is only one real solution (a repeated root).

Answer: The solution is $ x = -2 $.

Problem 9: Solving $ 3x^2 - 7x + 2 = 0 $

Step 1: Identify coefficients

• $ a = 3 $
• $ b = -7 $
• $ c = 2 $

Step 2: Apply the quadratic formula $ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)} = \frac{7 \pm \sqrt{49 - 24}}{6} $

Step 3: Simplify the discriminant $ 49 - 24 = 25 $, so: $ x = \frac{7 \pm 5}{6} $

Step 4: Solve for x

• $ x = \frac{7 + 5}{6} = \frac{12}{6} = 2 $
• $ x = \frac{7 - 5}{6} = \frac{2}{6} = \frac{1}{3} $

Answer: The solutions are $ x = 2 $ and $ x = \frac{1}{3} $.

Problem 10: Solving $ x^2 - 6x + 10 = 0 $

Step 1: Identify coefficients

• $ a = 1 $
• $ b = -6 $
• $ c = 10 $

Step 2: Apply the quadratic formula $ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)} = \frac{6 \pm \sqrt{36 - 40}}{2} $

Step 3: Simplify the discriminant $ 36 - 40 = -4 $, so: $ x = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} $

Step 4: Solve for x

• $ x = \frac{6 + 2i}{2} = 3 + i $
• $ x = \frac{6 - 2i}{2} = 3 - i $

Answer: The solutions are $ x = 3 + i $ and $ x = 3 - i $. These are complex conjugate roots.

Conclusion

These examples demonstrate the versatility of the quadratic formula. That said, we’ve seen it used to solve equations with two distinct real roots (Problems 6 & 7), a single repeated real root (Problem 8), and complex roots (Problem 10). The key to success lies in accurately identifying the coefficients a, b, and c, carefully substituting them into the formula, and meticulously simplifying the expression. Worth adding: the discriminant, $b^2 - 4ac$, provides valuable information about the nature of the roots: a positive discriminant indicates two distinct real roots, a zero discriminant indicates one repeated real root, and a negative discriminant indicates two complex conjugate roots. By mastering these steps, you can confidently tackle any quadratic equation and tap into a deeper understanding of algebraic problem-solving. Remember to always double-check your calculations, especially when dealing with negative signs and fractions, to ensure accurate results.