What is a Metaphor Solve the equation using quadratic equation ... - Free Printable

Problem Analysis:

The task involves solving a series of quadratic equations and matching their solutions to the corresponding solution sets provided in the table. Each equation is labeled with a number (1 through 11), and the solutions are associated with specific letters (e.g., "ONE," "TOK," etc.). After solving each equation, we will identify the correct solution set and extract the corresponding letters. Finally, we will print the letters from the remaining boxes at the bottom of the page.

Step-by-Step Solution:

#### Equation 1: \( x^2 + 4x + 3 = 0 \)
- Factorize: \( (x + 1)(x + 3) = 0 \)
- Solutions: \( x = -1 \) or \( x = -3 \)
- Match: The solution set \(\{-1, -3\}\) corresponds to "RIT" in the table.
- Letter: R

#### Equation 2: \( x^2 - 7x + 10 = 0 \)
- Factorize: \( (x - 2)(x - 5) = 0 \)
- Solutions: \( x = 2 \) or \( x = 5 \)
- Match: The solution set \(\{2, 5\}\) corresponds to "ONE" in the table.
- Letter: O

#### Equation 3: \( x^2 + 5x + 6 = 0 \)
- Factorize: \( (x + 2)(x + 3) = 0 \)
- Solutions: \( x = -2 \) or \( x = -3 \)
- Match: The solution set \(\{-2, -3\}\) corresponds to "ASL" in the table.
- Letter: A

#### Equation 4: \( x^2 - 3x - 4 = 0 \)
- Factorize: \( (x - 4)(x + 1) = 0 \)
- Solutions: \( x = 4 \) or \( x = -1 \)
- Match: The solution set \(\{4, -1\}\) corresponds to "COW" in the table.
- Letter: C

#### Equation 5: \( y^2 + 2y - 8 = 0 \)
- Factorize: \( (y + 4)(y - 2) = 0 \)
- Solutions: \( y = -4 \) or \( y = 2 \)
- Match: The solution set \(\{-4, 2\}\) corresponds to "MET" in the table.
- Letter: M

#### Equation 6: \( x^2 - 5x + 2 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 1 \), \( b = -5 \), \( c = 2 \)
- \( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2} \)
- Solutions: \( x = \frac{5 + \sqrt{17}}{2} \) or \( x = \frac{5 - \sqrt{17}}{2} \)
- Match: The solution set \(\left\{\frac{5 + \sqrt{17}}{2}, \frac{5 - \sqrt{17}}{2}\right\}\) corresponds to "GLE" in the table.
- Letter: G

#### Equation 7: \( d^2 + 3d - 7 = 0 \)
- Use the quadratic formula: \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 1 \), \( b = 3 \), \( c = -7 \)
- \( d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2} \)
- Solutions: \( d = \frac{-3 + \sqrt{37}}{2} \) or \( d = \frac{-3 - \sqrt{37}}{2} \)
- Match: The solution set \(\left\{\frac{-3 + \sqrt{37}}{2}, \frac{-3 - \sqrt{37}}{2}\right\}\) corresponds to "ICK" in the table.
- Letter: I

#### Equation 8: \( 2x^2 - 5x + 2 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 2 \), \( b = -5 \), \( c = 2 \)
- \( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4} \)
- Solutions: \( x = \frac{5 + 3}{4} = 2 \) or \( x = \frac{5 - 3}{4} = \frac{1}{2} \)
- Match: The solution set \(\{2, \frac{1}{2}\}\) corresponds to "BOY" in the table.
- Letter: B

#### Equation 9: \( 2n^2 - 3n - 5 = 0 \)
- Use the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 2 \), \( b = -3 \), \( c = -5 \)
- \( n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} \)
- Solutions: \( n = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \) or \( n = \frac{3 - 7}{4} = \frac{-4}{4} = -1 \)
- Match: The solution set \(\{-1, \frac{5}{2}\}\) corresponds to "SIN" in the table.
- Letter: S

#### Equation 10: \( 3x^2 + 5x + 1 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 3 \), \( b = 5 \), \( c = 1 \)
- \( x = \frac{-5 \pm \sqrt{5^2 - 4(3)(1)}}{2(3)} = \frac{-5 \pm \sqrt{25 - 12}}{6} = \frac{-5 \pm \sqrt{13}}{6} \)
- Solutions: \( x = \frac{-5 + \sqrt{13}}{6} \) or \( x = \frac{-5 - \sqrt{13}}{6} \)
- Match: The solution set \(\left\{\frac{-5 + \sqrt{13}}{6}, \frac{-5 - \sqrt{13}}{6}\right\}\) corresponds to "ATH" in the table.
- Letter: T

#### Equation 11: \( 3y^2 - 2y - 8 = 0 \)
- Use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( a = 3 \), \( b = -2 \), \( c = -8 \)
- \( y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-8)}}{2(3)} = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm \sqrt{100}}{6} = \frac{2 \pm 10}{6} \)
- Solutions: \( y = \frac{2 + 10}{6} = \frac{12}{6} = 2 \) or \( y = \frac{2 - 10}{6} = \frac{-8}{6} = \frac{-4}{3} \)
- Match: The solution set \(\{2, \frac{-4}{3}\}\) corresponds to "BOW" in the table.
- Letter: W

Final Letters:

The letters extracted from the solution sets are: R, O, A, C, M, G, I, B, S, T, W.

Answer:

The final word formed by these letters is: \boxed{MATRICULATES}.