5. Practice Questions

1. Solve using factorization: 𝑥² − 7𝑥 + 12 = 0

To solve the quadratic equation 𝑥² − 7𝑥 + 12 = 0 by factorization, we look for two numbers that multiply to +12 (the constant term) and add to -7 (the coefficient of the linear term, -7). The numbers -3 and -4 satisfy this condition, so we can express the quadratic as (𝑥 - 3)(𝑥 - 4) = 0. We then set each factor equal to zero: 𝑥 - 3 = 0 and 𝑥 - 4 = 0, resulting in solutions 𝑥 = 3 and 𝑥 = 4. Therefore, the roots of the equation are 3 and 4.

Think of this as finding two numbers that can form a rectangular area. If you need to create a garden area of 12 square meters, and you want the length and width to differ by 7 meters, you'll discover that a length of 4 meters and a width of 3 meters work perfectly, hence the numbers you are looking for.

2. Find the roots of: 3𝑥² + 5𝑥 − 2 = 0 using the quadratic formula.

To find the roots using the quadratic formula, we apply the formula x = (−𝑏 ± √(𝑏² − 4𝑎𝑐)) / (2𝑎), where a = 3, b = 5, and c = -2. First, we calculate the discriminant: 𝛥 = 𝑏² − 4𝑎𝑐 = 5² − 4(3)(-2) = 25 + 24 = 49. Since our discriminant is positive, it indicates there are two distinct real roots. Using the formula, we have x = [−5 ± √49] / [2(3)]. This simplifies to x = (−5 ± 7) / 6. Therefore, the two solutions are x = (2) / 6 = 1/3 and x = (−12) / 6 = −2.

Imagine you are throwing a ball vertically. By calculating the trajectory using the quadratic formula, you determine precisely when the ball reaches both its maximum height and when it will touch the ground again. These moments represent the roots of the quadratic equation.

3. Complete the square: 𝑥² + 8𝑥 + 10 = 0

To complete the square for the equation 𝑥² + 8𝑥 + 10 = 0, we start by isolating the constant on one side: 𝑥² + 8𝑥 = -10. Next, we take half of 8, square it, and add it to both sides: (8/2)² = 16. Thus, we rewrite it as 𝑥² + 8𝑥 + 16 = 6. This simplifies to (𝑥 + 4)² = 6. To solve for x, we take the square root of both sides: 𝑥 + 4 = ±√6, so 𝑥 = -4 ± √6, resulting in two solutions.

Completing the square is akin to adjusting the shape of a piece of dough to form a perfect circle. Just as you manipulate the dough to create an equal distance from the center point to the edges, in equations, you adjust the equation to manifest a perfect square.

4. Sketch the graph of: 𝑓(𝑥) = −𝑥² + 6𝑥 − 5, and find the vertex and axis of symmetry.

To graph the quadratic function 𝑓(𝑥) = −𝑥² + 6𝑥 − 5, we first identify that it's a downward-opening parabola because the coefficient of 𝑥² is negative. The next step is to find the vertex using the vertex formula 𝑥 = -b/2a. Here, a = -1 and b = 6, so 𝑥 = -6 / (2*-1) = 3. We then find the function value at this x-coordinate: 𝑓(3) = −(3)² + 6(3) − 5 = 4, giving us the vertex point (3, 4). The axis of symmetry is the vertical line x = 3. We can plot points on either side of the vertex to sketch the parabola.

Picture the trajectory of a fountain spray. The peak of the spray reaches the maximum height (the vertex), and as it falls back down, it's like the graph of a parabola reflecting its path. The axis of symmetry is the imaginary line that cuts through the center of this fountain's arc.

5. Word Problem: A ball is thrown upward with height modeled by: ℎ(𝑡) = −5𝑡²+20𝑡+1 a) Find the maximum height of the ball. b) After how many seconds does the ball hit the ground?

For the quadratic function ℎ(𝑡) = −5𝑡² + 20𝑡 + 1, the maximum height occurs at its vertex. We find the vertex using 𝑡 = -b/2a, giving us 𝑡 = 20 / (2*-5) = 2 seconds. Substituting this back into the height function: ℎ(2) = −5(2)² + 20(2) + 1 = 41 meters. To determine when the ball hits the ground, we set ℎ(𝑡) = 0 and solve: −5𝑡² + 20𝑡 + 1 = 0 using the quadratic formula. The solutions yield the time at which this occurs, which is when the ball returns to a height of 0.

Imagine you're at a sporting event watching someone throw a ball into the air. At first, it's ascending quickly to its peak height (maximum height) and then starts to come back down. By calculating these vital points, you can tell precisely how high the ball will go and when it will hit the ground, just like the modeling done in this problem.

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