Unit 4 Solving Quadratic Equations Homework 9 Answer Key

Unit 4: Solving Quadratic Equations – Homework 9 Answer Key

Quadratic equations appear everywhere—from calculating projectile paths to designing parabolic arches. Below you’ll find a comprehensive answer key for each problem, along with detailed explanations that clarify the reasoning behind every step. Practically speaking, homework 9 is a staple assignment that tests mastery of these techniques. In Unit 4 of most algebra courses, students learn the three main methods for solving them: factoring, completing the square, and the quadratic formula. Use this guide to double‑check your work, understand common pitfalls, and reinforce the concepts that will help you ace future exams.

1. Problem 1 – Factoring

Equation:
(x^2 - 5x + 6 = 0)

Solution Steps:

1. Identify two numbers whose product equals the constant term (+6) and whose sum equals the coefficient of (x) (‑5).
   The pair is (‑2, ‑3) because ((-2) \times (-3) = 6) and ((-2) + (-3) = -5).

2. Rewrite the quadratic using these numbers:
   ((x - 2)(x - 3) = 0).

3. Apply the Zero‑Product Property:
   Set each factor equal to zero.

   • (x - 2 = 0 \Rightarrow x = 2)
   • (x - 3 = 0 \Rightarrow x = 3)

Answer: (x = 2) or (x = 3)

2. Problem 2 – Completing the Square

Equation:
(x^2 + 4x - 5 = 0)

Solution Steps:

1. Move the constant term to the right side:
   (x^2 + 4x = 5).

2. Add the square of half the coefficient of (x) to both sides:
   (\left(\frac{4}{2}\right)^2 = 4).
   (x^2 + 4x + 4 = 5 + 4).

3. Rewrite the left side as a perfect square:
   ((x + 2)^2 = 9).

4. Take the square root of both sides, remembering the ± sign:
   (x + 2 = \pm 3).

5. Solve for (x):

   • (x + 2 = 3 \Rightarrow x = 1)
   • (x + 2 = -3 \Rightarrow x = -5)

Answer: (x = 1) or (x = -5)

3. Problem 3 – Quadratic Formula

Equation:
(2x^2 + 7x - 3 = 0)

Solution Steps:

1. Identify (a = 2), (b = 7), (c = -3).

2. Compute the discriminant (\Delta = b^2 - 4ac):
   (\Delta = 7^2 - 4(2)(-3) = 49 + 24 = 73).

3. Apply the quadratic formula:
   (x = \frac{-b \pm \sqrt{\Delta}}{2a}).
   (x = \frac{-7 \pm \sqrt{73}}{4}).

4. Simplify (if desired, keep the radical form):
   (x = \frac{-7 + \sqrt{73}}{4}) or (x = \frac{-7 - \sqrt{73}}{4}).

Answer: (x = \frac{-7 + \sqrt{73}}{4}) or (x = \frac{-7 - \sqrt{73}}{4})

4. Problem 4 – Factoring with a Leading Coefficient

Equation:
(3x^2 - 10x + 4 = 0)

Solution Steps:

1. Multiply (a) and (c): (3 \times 4 = 12).

2. Find two numbers whose product is 12 and sum is (-10):
   The pair is (-6) and (-4).

3. Rewrite the middle term using these numbers:
   (3x^2 - 6x - 4x + 4 = 0).

4. Factor by grouping:
   ((3x^2 - 6x) + (-4x + 4) = 3x(x - 2) - 2(x - 2) = (3x - 2)(x - 2) = 0).

5. Solve for (x):

   • (3x - 2 = 0 \Rightarrow x = \frac{2}{3})
   • (x - 2 = 0 \Rightarrow x = 2)

Answer: (x = \frac{2}{3}) or (x = 2)

5. Problem 5 – Completing the Square with a Leading Coefficient

Equation:
(4x^2 + 8x - 12 = 0)

Solution Steps:

1. Divide the entire equation by 4 to simplify:
   (x^2 + 2x - 3 = 0).

2. Move the constant term:
   (x^2 + 2x = 3).

3. Add ((\frac{2}{2})^2 = 1) to both sides:
   (x^2 + 2x + 1 = 4).

4. Rewrite as a perfect square:
   ((x + 1)^2 = 4).

5. Take the square root:
   (x + 1 = \pm 2).

6. Solve for (x):

   • (x + 1 = 2 \Rightarrow x = 1)
   • (x + 1 = -2 \Rightarrow x = -3)

Answer: (x = 1) or (x = -3)

6. Problem 6 – Quadratic Formula with Negative Discriminant

Equation:
(x^2 - 4x + 5 = 0)

Solution Steps:

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1. Identify (a = 1), (b = -4), (c = 5).

2. Compute the discriminant:
   (\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4).

3. Apply the quadratic formula:
   (x = \frac{4 \pm \sqrt{-4}}{2}).

4. Simplify the radical:
   (\sqrt{-4} = 2i) (where (i) is the imaginary unit).

5. Finalize the complex roots:
   (x = \frac{4 \pm 2i}{2} = 2 \pm i).

Answer: (x = 2 + i) or (x = 2 - i)

7. Problem 7 – Factoring with a Non‑Integer Root

Equation:
(x^2 - 6x + 9 = 0)

Solution Steps:

1. Recognize a perfect square:
   (x^2 - 6x + 9 = (x - 3)^2).

2. Set the factor equal to zero:
   (x - 3 = 0 \Rightarrow x = 3).

3. Since the factor is squared, the root is repeated (a double root).

Answer: (x = 3) (double root)

8. Problem 8 – Using the Quadratic Formula with Large Coefficients

Equation:
(5x^2 + 11x - 8 = 0)

Solution Steps:

1. Identify (a = 5), (b = 11), (c = -8).

2. Compute the discriminant:
   (\Delta = 11^2 - 4(5)(-8) = 121 + 160 = 281).

3. Apply the quadratic formula:
   (x = \frac{-11 \pm \sqrt{281}}{10}).

4. Simplify:
   (x = \frac{-11 + \sqrt{281}}{10}) or (x = \frac{-11 - \sqrt{281}}{10}).

Answer: (x = \frac{-11 + \sqrt{281}}{10}) or (x = \frac{-11 - \sqrt{281}}{10})

9. Problem 9 – Completing the Square with a Fractional Coefficient

Equation:
(\frac{1}{2}x^2 + 3x - 4 = 0)

Solution Steps:

1. Multiply by 2 to clear the fraction:
   (x^2 + 6x - 8 = 0).

2. Move the constant term:
   (x^2 + 6x = 8).

3. Add ((\frac{6}{2})^2 = 9):
   (x^2 + 6x + 9 = 17).

4. Rewrite as a perfect square:
   ((x + 3)^2 = 17).

5. Take the square root:
   (x + 3 = \pm \sqrt{17}).

6. Solve for (x):

   • (x = -3 + \sqrt{17})
   • (x = -3 - \sqrt{17})

Answer: (x = -3 + \sqrt{17}) or (x = -3 - \sqrt{17})

10. Problem 10 – Verifying a Solution

Equation:
(x^2 - 2x - 3 = 0)

Given Solution: (x = 3)

Verification Steps:

1. Substitute (x = 3) into the left‑hand side:
   (3^2 - 2(3) - 3 = 9 - 6 - 3 = 0).

2. Result equals zero, confirming that (x = 3) is indeed a root.

Answer: Verified; (x = 3) satisfies the equation.

Frequently Asked Questions (FAQ)

Q1: When should I use factoring instead of the quadratic formula?

A: Factoring is quickest when the quadratic can be expressed as a product of two binomials with integer coefficients. If the discriminant is a perfect square, factoring is usually the most efficient.

Q2: What if the discriminant is negative?

A: A negative discriminant indicates no real solutions; instead, the quadratic has two complex conjugate roots. Use the quadratic formula to express them in terms of the imaginary unit (i).

Q3: How do I know if a quadratic is a perfect square?

A: Check if the constant term is the square of half the middle coefficient. To give you an idea, (x^2 - 6x + 9) is ((x - 3)^2) because ((\frac{-6}{2})^2 = 9).

Q4: Can I use completing the square when the leading coefficient is not 1?

A: Yes, but first factor out the leading coefficient from the (x^2) and (x) terms before completing the square. This ensures the left side becomes a perfect square.

Q5: Why do some quadratics have a repeated root?

A: A repeated root occurs when the discriminant equals zero. The quadratic can then be written as ((x - r)^2 = 0), giving a single solution (x = r) with multiplicity two.

Conclusion

Mastering the three core techniques—factoring, completing the square, and the quadratic formula—provides a dependable toolkit for tackling any quadratic equation. Homework 9 serves as an excellent practice ground, reinforcing the logic behind each method and highlighting common pitfalls such as sign errors or overlooking complex solutions. Because of that, by following the step‑by‑step solutions above, you can confidently verify your answers, gain deeper insight into the structure of quadratics, and prepare for more advanced algebraic concepts. Keep practicing, and soon solving quadratic equations will feel as natural as solving for a missing variable in everyday life.