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Quadratic Equations

Solve:

1. (4s + 9)^2 = 36
2. f(x) = x^2 + 18x
3. z^2 + 18z + 64 = 0
5. 7n^2 = 10n - 2
6. 5r^2 + 20r = -18
7. -7x^2 - 5x = 9
8. f(x) = 3x^2 - 5x - 1

Find the vertex: 13. f(x) = (x + 6)^2 - 2

Graph:

12. f(x) = -4(x + 7)^2 + 4
15. f(x) = -x^2 + 2x - 7

Find the x and y intercepts: 16. f(x) = 2x^2 + 6x + 1

Six word problems (problems 4., 9., and 17. - 20.)

Solve for r: 10. A = (1/3)(pi)r^2

Find the x-intercepts: 11. f(x) = (x^2 - 13x)^2 + 25(x^2 - 13x) - 150

Find the maximum or minimum: 14. f(x) = 5x^2 - 80x + 327

See attached file for full problem description.

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Solution Preview

Most of these problems can be solved using algebra (as opposed to geometry). Like in previous homework assignments, take square roots, factor, complete the square, or use the quadratic equation to find the solution. For this homework assignment, the geometry part is understanding parabolas. Make sure you can complete the square to put an equation into the form y = a(x - h)2 + k, and know that the vertex of a parabola is the minimum or the maximum (depending on whether the parabola opens up or down).

Notice that both sides of the equation are perfect squares. Therefore, you can simply take the square root of both sides, then solve for s. Remember that √36 can equal 6 and -6.

(4s + 9)2 = 36
4s + 9 = 6 4s + 9 = -6
4s = -3 4s = -15
s = -3/4 s = -15/4

s = -15/4, -3/4

The x-intercepts occur when f(x) = 0. Set f(x) = 0, then solve for x. The easiest method in this case is factoring (factor the x out of both terms on the right hand side).

f(x) = x2 + 18x
0 = x2 + 18x
0 = x(x + 18)
x = 0 OR x + 18 = 0
x = -18

The x-intercepts are (0, 0) and (-18, 0).

When you complete the square, you add a term so that z2 + 18z + the added term is a perfect square. You should add a 17. I got this by dividing the 18 by 2, then squaring it to see that we want the left side to look like z2 + 18z + 81 which equals (z + 9)2 ... To get the 81 on the left, you need to add 17 to 64). Since you add 17 to the left, you also need to add it to the right of the equation.

z2 + 18z + 64 = 0
z2 + 18z + 17 + 64 = 17
z2 + 18z + 81 = 17
(z + 9)(z + 9) = 17
(z + 9)2 = 17

z + 9 = √17 z + 9 = -√17
z = √17 - 9 z = -√17 - 9

z = -(√17 + 9), √17 - 9

First, put everything on one side of the equation. Then you can use the quadratic equation to solve for n. Remember than the quadratic equation always works, but you have to have all the terms on one side of the equation and have the other side of the equation equal to 0.

7n2 = -10n -2
7n2 + 10n + 2 = 0

n = -10 ± √(100 - 4*7*2)
14

n = -10 ± √(44) = -10 ± 2√(11)
14 14

n = -5 ± √(11)
7

You can solve this in the same way as question 5.

5r2 + 20r = -18
5r2 + 20r + 18 = 0

r = -20 ± √(400 - 4*5*18)
10

r = -20 ± √(40) = -20 ± 2√(10)
10 10

r = -10 ± √(10)
5

-7x2 ...

Solution Summary

The solution includes answers and complete, step-by-step explanations to twenty algebra problems involving quadratic equations and parabolas. These questions include:

solving equations,
finding x- and y-intercepts,
completing the square,
finding the vertex of a parabola,
graphing,
word problems, and
finding the maximum or minimum value of a function.

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