Unit 4: Solving Quadratic Equations – Homework 9 Answer Key
Quadratic equations appear everywhere—from calculating projectile paths to designing parabolic arches. Homework 9 is a staple assignment that tests mastery of these techniques. Plus, in Unit 4 of most algebra courses, students learn the three main methods for solving them: factoring, completing the square, and the quadratic formula. Consider this: below you’ll find a comprehensive answer key for each problem, along with detailed explanations that clarify the reasoning behind every step. 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:
-
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) Small thing, real impact.. -
Rewrite the quadratic using these numbers:
((x - 2)(x - 3) = 0). -
Apply the Zero‑Product Property:
Set each factor equal to zero Practical, not theoretical..- (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:
-
Move the constant term to the right side:
(x^2 + 4x = 5) It's one of those things that adds up. No workaround needed.. -
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) Worth keeping that in mind.. -
Rewrite the left side as a perfect square:
((x + 2)^2 = 9). -
Take the square root of both sides, remembering the ± sign:
(x + 2 = \pm 3). -
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:
-
Identify (a = 2), (b = 7), (c = -3).
-
Compute the discriminant (\Delta = b^2 - 4ac):
(\Delta = 7^2 - 4(2)(-3) = 49 + 24 = 73) Worth keeping that in mind.. -
Apply the quadratic formula:
(x = \frac{-b \pm \sqrt{\Delta}}{2a}).
(x = \frac{-7 \pm \sqrt{73}}{4}). -
Simplify (if desired, keep the radical form):
(x = \frac{-7 + \sqrt{73}}{4}) or (x = \frac{-7 - \sqrt{73}}{4}) Easy to understand, harder to ignore..
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:
-
Multiply (a) and (c): (3 \times 4 = 12).
-
Find two numbers whose product is 12 and sum is (-10):
The pair is (-6) and (-4). -
Rewrite the middle term using these numbers:
(3x^2 - 6x - 4x + 4 = 0). -
Factor by grouping:
((3x^2 - 6x) + (-4x + 4) = 3x(x - 2) - 2(x - 2) = (3x - 2)(x - 2) = 0) Easy to understand, harder to ignore.. -
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:
-
Divide the entire equation by 4 to simplify:
(x^2 + 2x - 3 = 0) And that's really what it comes down to.. -
Move the constant term:
(x^2 + 2x = 3). -
Add ((\frac{2}{2})^2 = 1) to both sides:
(x^2 + 2x + 1 = 4) Small thing, real impact. Simple as that.. -
Rewrite as a perfect square:
((x + 1)^2 = 4). -
Take the square root:
(x + 1 = \pm 2) Simple, but easy to overlook.. -
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:
-
Identify (a = 1), (b = -4), (c = 5).
-
Compute the discriminant:
(\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4). -
Apply the quadratic formula:
(x = \frac{4 \pm \sqrt{-4}}{2}) Took long enough.. -
Simplify the radical:
(\sqrt{-4} = 2i) (where (i) is the imaginary unit). -
Finalize the complex roots:
(x = \frac{4 \pm 2i}{2} = 2 \pm i) It's one of those things that adds up..
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:
-
Recognize a perfect square:
(x^2 - 6x + 9 = (x - 3)^2) Simple as that.. -
Set the factor equal to zero:
(x - 3 = 0 \Rightarrow x = 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:
-
Identify (a = 5), (b = 11), (c = -8).
-
Compute the discriminant:
(\Delta = 11^2 - 4(5)(-8) = 121 + 160 = 281). -
Apply the quadratic formula:
(x = \frac{-11 \pm \sqrt{281}}{10}) It's one of those things that adds up.. -
Simplify:
(x = \frac{-11 + \sqrt{281}}{10}) or (x = \frac{-11 - \sqrt{281}}{10}) Most people skip this — try not to. Worth knowing..
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:
-
Multiply by 2 to clear the fraction:
(x^2 + 6x - 8 = 0) Which is the point.. -
Move the constant term:
(x^2 + 6x = 8). -
Add ((\frac{6}{2})^2 = 9):
(x^2 + 6x + 9 = 17). -
Rewrite as a perfect square:
((x + 3)^2 = 17). -
Take the square root:
(x + 3 = \pm \sqrt{17}). -
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:
-
Substitute (x = 3) into the left‑hand side:
(3^2 - 2(3) - 3 = 9 - 6 - 3 = 0). -
Result equals zero, confirming that (x = 3) is indeed a root.
Answer: Verified; (x = 3) satisfies the equation Still holds up..
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) And that's really what it comes down to. Surprisingly effective..
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) And that's really what it comes down to..
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 That alone is useful..
Conclusion
Mastering the three core techniques—factoring, completing the square, and the quadratic formula—provides a reliable 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. 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 Surprisingly effective..
