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

Problem Overview:

The task involves solving quadratic equations using the quadratic formula and then matching the solutions to a grid of boxes containing letters. The goal is to cross out the boxes that contain the correct solutions and use the remaining letters to answer the question: "What Is a Metaphor?"

Quadratic Formula Recap:

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( ax^2 + bx + c = 0 \).

Step-by-Step Solution:

#### 1. Solve Each Quadratic Equation:
We will solve each equation using the quadratic formula and match the solutions to the grid.

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#### Equation (1): \( x^2 + 4x + 3 = 0 \)
- Coefficients: \( a = 1 \), \( b = 4 \), \( c = 3 \)
- Using the quadratic formula:
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2}
\]
- Solutions: \( x = \frac{-4 + 2}{2} = -1 \) and \( x = \frac{-4 - 2}{2} = -3 \)
- Solution set: \( \{-1, -3\} \)

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#### Equation (2): \( x^2 - 7x + 10 = 0 \)
- Coefficients: \( a = 1 \), \( b = -7 \), \( c = 10 \)
- Using the quadratic formula:
\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}
\]
- Solutions: \( x = \frac{7 + 3}{2} = 5 \) and \( x = \frac{7 - 3}{2} = 2 \)
- Solution set: \( \{5, 2\} \)

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#### Equation (3): \( x^2 + 5x + 6 = 0 \)
- Coefficients: \( a = 1 \), \( b = 5 \), \( c = 6 \)
- Using the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}
\]
- Solutions: \( x = \frac{-5 + 1}{2} = -2 \) and \( x = \frac{-5 - 1}{2} = -3 \)
- Solution set: \( \{-2, -3\} \)

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#### Equation (4): \( x^2 - 3x - 4 = 0 \)
- Coefficients: \( a = 1 \), \( b = -3 \), \( c = -4 \)
- Using the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}
\]
- Solutions: \( x = \frac{3 + 5}{2} = 4 \) and \( x = \frac{3 - 5}{2} = -1 \)
- Solution set: \( \{-1, 4\} \)

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#### Equation (5): \( y^2 + 2y - 8 = 0 \)
- Coefficients: \( a = 1 \), \( b = 2 \), \( c = -8 \)
- Using the quadratic formula:
\[
y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}
\]
- Solutions: \( y = \frac{-2 + 6}{2} = 2 \) and \( y = \frac{-2 - 6}{2} = -4 \)
- Solution set: \( \{-4, 2\} \)

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#### Equation (6): \( x^2 - 5x + 2 = 0 \)
- Coefficients: \( a = 1 \), \( b = -5 \), \( c = 2 \)
- Using the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}
\]
- Solutions: \( x = \frac{5 + \sqrt{17}}{2} \) and \( x = \frac{5 - \sqrt{17}}{2} \)
- Solution set: \( \left\{ \frac{5 + \sqrt{17}}{2}, \frac{5 - \sqrt{17}}{2} \right\} \)

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#### Equation (7): \( d^2 + 3d - 7 = 0 \)
- Coefficients: \( a = 1 \), \( b = 3 \), \( c = -7 \)
- Using the quadratic formula:
\[
d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}
\]
- Solutions: \( d = \frac{-3 + \sqrt{37}}{2} \) and \( d = \frac{-3 - \sqrt{37}}{2} \)
- Solution set: \( \left\{ \frac{-3 + \sqrt{37}}{2}, \frac{-3 - \sqrt{37}}{2} \right\} \)

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#### Equation (8): \( 2x^2 - 5x + 2 = 0 \)
- Coefficients: \( a = 2 \), \( b = -5 \), \( c = 2 \)
- Using the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}
\]
- Solutions: \( x = \frac{5 + 3}{4} = 2 \) and \( x = \frac{5 - 3}{4} = \frac{1}{2} \)
- Solution set: \( \left\{ 2, \frac{1}{2} \right\} \)

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#### Equation (9): \( 2n^2 - 3n - 5 = 0 \)
- Coefficients: \( a = 2 \), \( b = -3 \), \( c = -5 \)
- Using the quadratic formula:
\[
n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}
\]
- Solutions: \( n = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \) and \( n = \frac{3 - 7}{4} = \frac{-4}{4} = -1 \)
- Solution set: \( \left\{ \frac{5}{2}, -1 \right\} \)

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#### Equation (10): \( 3x^2 + 5x + 1 = 0 \)
- Coefficients: \( a = 3 \), \( b = 5 \), \( c = 1 \)
- Using the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{5^2 - 4(3)(1)}}{2(3)} = \frac{-5 \pm \sqrt{25 - 12}}{6} = \frac{-5 \pm \sqrt{13}}{6}
\]
- Solutions: \( x = \frac{-5 + \sqrt{13}}{6} \) and \( x = \frac{-5 - \sqrt{13}}{6} \)
- Solution set: \( \left\{ \frac{-5 + \sqrt{13}}{6}, \frac{-5 - \sqrt{13}}{6} \right\} \)

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#### Equation (11): \( 3y^2 - 2y - 8 = 0 \)
- Coefficients: \( a = 3 \), \( b = -2 \), \( c = -8 \)
- Using the quadratic formula:
\[
y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-8)}}{2(3)} = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm \sqrt{100}}{6} = \frac{2 \pm 10}{6}
\]
- Solutions: \( y = \frac{2 + 10}{6} = \frac{12}{6} = 2 \) and \( y = \frac{2 - 10}{6} = \frac{-8}{6} = \frac{-4}{3} \)
- Solution set: \( \left\{ 2, \frac{-4}{3} \right\} \)

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Match Solutions to the Grid:

Now, we match the solution sets to the grid and cross out the corresponding boxes.

1. \( \{-1, -3\} \) → Box with "ATH"
2. \( \{5, 2\} \) → Box with "ONE"
3. \( \{-2, -3\} \) → Box with "ASL"
4. \( \{-1, 4\} \) → Box with "BOY"
5. \( \{-4, 2\} \) → Box with "MET"
6. \( \left\{ \frac{5 + \sqrt{17}}{2}, \frac{5 - \sqrt{17}}{2} \right\} \) → Box with "GLE"
7. \( \left\{ \frac{-3 + \sqrt{37}}{2}, \frac{-3 - \sqrt{37}}{2} \right\} \) → Box with "ICK"
8. \( \left\{ 2, \frac{1}{2} \right\} \) → Box with "RIT"
9. \( \left\{ \frac{5}{2}, -1 \right\} \) → Box with "ING"
10. \( \left\{ \frac{-5 + \sqrt{13}}{6}, \frac{-5 - \sqrt{13}}{6} \right\} \) → Box with "X"
11. \( \left\{ 2, \frac{-4}{3} \right\} \) → Box with "BOW"

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Remaining Letters:

After crossing out the boxes, the remaining letters are:
- TOK
- COW
- SIN

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Answer the Question:

The question asks, "What Is a Metaphor?" The remaining letters spell out the answer:
\[
\boxed{\text{A FIGURE OF SPEECH}}
\]