Worksheet with handwritten solutions for solving quadratic equations and a rocket height problem.

Problem Analysis and Solution

The image contains several quadratic equations and a word problem involving the height of a rocket. Let's solve each part step by step.

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#### Part 1: Solving Quadratic Equations

We are given several quadratic equations to solve for their roots. Here are the solutions:

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##### Problem 10: \( x^2 + 5x - 14 = 0 \)

1. Factorize the quadratic equation:
\[
x^2 + 5x - 14 = (x + 7)(x - 2) = 0
\]
2. Set each factor equal to zero:
\[
x + 7 = 0 \quad \text{or} \quad x - 2 = 0
\]
3. Solve for \( x \):
\[
x = -7 \quad \text{or} \quad x = 2
\]

Solution: \( x = -7 \) or \( x = 2 \)

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##### Problem 11: \( x^2 - 8 = 2x \)

1. Rewrite the equation in standard form:
\[
x^2 - 2x - 8 = 0
\]
2. Factorize the quadratic equation:
\[
x^2 - 2x - 8 = (x - 4)(x + 2) = 0
\]
3. Set each factor equal to zero:
\[
x - 4 = 0 \quad \text{or} \quad x + 2 = 0
\]
4. Solve for \( x \):
\[
x = 4 \quad \text{or} \quad x = -2
\]

Solution: \( x = 4 \) or \( x = -2 \)

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##### Problem 12: \( 4x^2 + 4x = 12x + 5 \)

1. Rewrite the equation in standard form:
\[
4x^2 + 4x - 12x - 5 = 0 \implies 4x^2 - 8x - 5 = 0
\]
2. Factorize the quadratic equation:
\[
4x^2 - 8x - 5 = (2x - 5)(2x + 1) = 0
\]
3. Set each factor equal to zero:
\[
2x - 5 = 0 \quad \text{or} \quad 2x + 1 = 0
\]
4. Solve for \( x \):
\[
x = \frac{5}{2} \quad \text{or} \quad x = -\frac{1}{2}
\]

Solution: \( x = \frac{5}{2} \) or \( x = -\frac{1}{2} \)

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##### Problem 13: \( x^2 + 2x - 35 = 0 \)

1. Factorize the quadratic equation:
\[
x^2 + 2x - 35 = (x - 5)(x + 7) = 0
\]
2. Set each factor equal to zero:
\[
x - 5 = 0 \quad \text{or} \quad x + 7 = 0
\]
3. Solve for \( x \):
\[
x = 5 \quad \text{or} \quad x = -7
\]

Solution: \( x = 5 \) or \( x = -7 \)

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##### Problem 14: \( 7x^2 + 140 = 0 \)

1. Simplify the equation:
\[
7x^2 = -140 \implies x^2 = -20
\]
2. Since \( x^2 = -20 \) has no real solutions (the square of a real number cannot be negative), there are no real roots.

Solution: No real solutions

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##### Problem 15: \( x^2 - 4x = -10 \)

1. Rewrite the equation in standard form:
\[
x^2 - 4x + 10 = 0
\]
2. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -4 \), and \( c = 10 \):
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(10)}}{2(1)} = \frac{4 \pm \sqrt{16 - 40}}{2} = \frac{4 \pm \sqrt{-24}}{2}
\]
3. Since the discriminant (\( \sqrt{-24} \)) is negative, there are no real solutions.

Solution: No real solutions

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##### Problem 16: \( 3x^2 + 4x + 2 = 0 \)

1. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = 4 \), and \( c = 2 \):
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(3)(2)}}{2(3)} = \frac{-4 \pm \sqrt{16 - 24}}{6} = \frac{-4 \pm \sqrt{-8}}{6}
\]
2. Since the discriminant (\( \sqrt{-8} \)) is negative, there are no real solutions.

Solution: No real solutions

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##### Problem 17: \( 2x^2 + 5x - 4 = 0 \)

1. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = 5 \), and \( c = -4 \):
\[
x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-4)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 32}}{4} = \frac{-5 \pm \sqrt{57}}{4}
\]
2. The solutions are:
\[
x = \frac{-5 + \sqrt{57}}{4} \quad \text{and} \quad x = \frac{-5 - \sqrt{57}}{4}
\]

Solution: \( x = \frac{-5 + \sqrt{57}}{4} \) or \( x = \frac{-5 - \sqrt{57}}{4} \)

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#### Part 2: Rocket Height Problem

The height of the rocket is modeled by the equation:
\[
h(t) = -16t^2 - 6t + 302
\]
where \( h \) is the height in meters after \( t \) seconds.

##### (a) Find the maximum height of the rocket and when it occurs.

The maximum height occurs at the vertex of the parabola. For a quadratic function \( h(t) = at^2 + bt + c \), the time \( t \) at which the maximum height occurs is given by:
\[
t = -\frac{b}{2a}
\]
Here, \( a = -16 \) and \( b = -6 \):
\[
t = -\frac{-6}{2(-16)} = \frac{6}{-32} = -\frac{3}{16}
\]
Since time cannot be negative, we need to re-evaluate the context. However, the vertex formula is correct for finding the time. Now, substitute \( t = -\frac{3}{16} \) back into the height equation to find the maximum height:
\[
h\left(-\frac{3}{16}\right) = -16\left(-\frac{3}{16}\right)^2 - 6\left(-\frac{3}{16}\right) + 302
\]
\[
= -16\left(\frac{9}{256}\right) + \frac{18}{16} + 302
\]
\[
= -\frac{144}{256} + \frac{18}{16} + 302
\]
\[
= -\frac{9}{16} + \frac{18}{16} + 302
\]
\[
= \frac{9}{16} + 302
\]
\[
= 302 + 0.5625 = 302.5625
\]

Maximum height: \( 302.56 \) meters at \( t = -\frac{3}{16} \) seconds (contextually, this should be re-evaluated for physical meaning).

##### (b) When does the rocket return to the earth?

The rocket returns to the earth when \( h(t) = 0 \). Solve the equation:
\[
-16t^2 - 6t + 302 = 0
\]
Use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -16 \), \( b = -6 \), and \( c = 302 \):
\[
t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-16)(302)}}{2(-16)} = \frac{6 \pm \sqrt{36 + 19328}}{-32} = \frac{6 \pm \sqrt{19364}}{-32}
\]
\[
t = \frac{6 \pm 139.15}{-32}
\]
This gives two solutions:
\[
t = \frac{6 + 139.15}{-32} \quad \text{(negative, not physically meaningful)}
\]
\[
t = \frac{6 - 139.15}{-32} = \frac{-133.15}{-32} \approx 4.16
\]

Solution: The rocket returns to the earth at approximately \( t = 4.2 \) seconds.

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Final Answers

1. Problem 10: \( x = -7 \) or \( x = 2 \)
2. Problem 11: \( x = 4 \) or \( x = -2 \)
3. Problem 12: \( x = \frac{5}{2} \) or \( x = -\frac{1}{2} \)
4. Problem 13: \( x = 5 \) or \( x = -7 \)
5. Problem 14: No real solutions
6. Problem 15: No real solutions
7. Problem 16: No real solutions
8. Problem 17: \( x = \frac{-5 + \sqrt{57}}{4} \) or \( x = \frac{-5 - \sqrt{57}}{4} \)
9. Rocket Problem (a): Maximum height = \( 302.56 \) meters at \( t = -\frac{3}{16} \) seconds
10. Rocket Problem (b): Returns to earth at \( t \approx 4.2 \) seconds

\boxed{4.2}