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Twenty Algebra Questions

Algebra Questions. See attached file for full problem description.

1. Solve 15x = 2x2 + 16
2. The width is 7 feet less than the length; the area is 18 square feet. Find the length and width.
3. Solve x2 + 4x - 12 < 0
4. Solve (x + 3)/(x - 4) < 0
5. Find f(-13) for f(x) = third root of (2x - 1)
6. Rewrite (third root of (7xy))4
7. Simplify fifth root of (p14q9r23)
8. Solve third root of (3y + 6) + 2 = 5
9. Solve for t: L = (Mt + g)/t
10. Graph f(x) = |x| - 4
11. Graph f(x) = x2 - 1
12. Simplify (x3y/pq2)-3
13. Subtract and simplify 4/(5a2 - 5a) - 2/(5a - 5)
14. Find the domain of f(x) = 15/(3x - 8)
15. - 18. See file for description.
19. Solve 3(r - 6) + 2 > 4(r + 2) - 21
20. Graph 2x + 3y >= 6

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Solution Preview

1. Solve 15x = 2x2 + 16

Put everything on one side of the equation, then use the quadratic formula:

15x = 2x2 + 16
0 = 2x2 - 15x + 16
x = 15 ±&#8730;(152 - 4(2)(16))
2(2)
x = 15 ±&#8730;97
4
x = 1.29, 6.21

2. The width is 7 feet less than the length; the area is 18 square feet. Find the length and width.

Let x be the length of the flowerbed. Then, x - 7 would be the width. Since, area is equal to length x width, we have the following equation:

(x)(x - 7) = 18

Multiply, put everything on one side of the equation, then factor to solve for x:

(x)(x - 7) = 18
x2 - 7x = 18
x2 - 7x -18 = 0
(x - 9)(x + 2) = 0
x = 9, -2

The only possible answer for x is x = 9 (a length can't be negative). Therefore, the sides are 9 and 2.

3. Solve x2 + 4x - 12 < 0

x2 + 4x - 12 < 0
(x + 6)(x - 2) < 0

This happens when x is between -6 and 2. In solution set notation, this is {x | -6 < x < 2}; in interval notation, this is (-6, 2). ...

Solution Summary

This problem set has twenty questions involving solving equations, simplifying expressions, exponents, roots, absolute value, graphing, and word problems.

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