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Math 121, Test 1, 23 January 2004
Name: ANSWERS
Instructions. Do all of the following 14 problems. Each problem is worth 5pts. Show all appropriate details in your solutions. Calculators are not allowed. Good Luck!
- Write the expression |x โ 5 | + | 2 x + 2| given 0 < x < 4 without absolute value symbols.
Answer. |x โ 5 | + | 2 x + 2| = โ(x โ 5) + 2x + 2 = x + 7 if 0 < x < 4.
- Simplify the complex fraction
1 x โ^
3 x+ 2 + (^) x+1^5
Answer. Simplify by multiplying the numerator and denominator by the LCD x(x + 1):
1 x โ^
3 x+ 2 + (^) x+1^5 =
x โ^
3 x+
x(x + 1) ( 2 + (^) x+1^5
x(x + 1) =^
x + 1 โ 3 x 2 x(x + 1) + 5x =^
1 โ 2 x x(2x + 7).
- Factor the expression 2x + 1 โ 4 xy โ 2 y.
Answer. 2 x + 1 โ 4 xy โ 2 y = 1(2x + 1) โ 2 y(2x + 1) = (1 โ 2 y)(2x + 1).
- Simplify: (2a
โ (^1) bcโ (^2) ) 2 (3โ^1 b)(2โ^1 acโ^2 )^3
Answer. (2a
โ (^1) bcโ (^2) ) 2 (3โ^1 b)(2โ^1 acโ^2 )^3 =^
4 aโ^2 b^2 cโ^4 3 โ^12 โ^3 a^3 bcโ^6 =
96 bc^2 a^5.
- Simplify: (^) x (^2) โx^ โ 6 x^2 + 9 รท x
x^2 โ x โ 6.
Answer. x^ โ^2 x^2 โ 6 x + 9
รท x
x^2 โ x โ 6
= x^ โ^2 (x โ 3)(x โ 3)
ยท (x^ โ^ 3)(x^ + 2) (x + 2)(x โ 2)
x โ 3
- Write the complex number^4 โ^3 i 3 + 2i
in standard form.
Answer.
4 โ 3 i 3 + 2i =
(4 โ 3 i)(3 โ 2 i) (3 + 2i)(3 โ 2 i) =
6 โ 17 i 13 =^
13 โ^
13 i
- Solve the equation^3 x^ + 7 2 x โ 3
Answer. Multiply both sides of the equation by (2x โ 3) to obtain: 3x + 7 = โ2(2x โ 3), or
3 x + 7 = โ 4 x + 6. Therefore, 7x = โ1, and so x = โ^17.
- Pump 1 can drain a pool in 6 hours, while pump 2 can drain the same pool in 7 hours. How long would it take to drain the pool using both pumps?
Answer. Let t be the amount of time in hours it would take both pumps to drain the pool. Then (^) t
7 +^
t 6 = 1
and so multiplying both sides of this equation by 42 yields 6t + 7t = 42. Therefore, t =^42 13 hours, or 3 133 hours.
- Solve the quadratic equation^13 x^2 +^16 x + 2 = 0 by using the quadratic formula.
Answer. Multiply both sides of this equation by 6 to eliminate the fractions. Therefore, we solve 2 x^2 + x + 12 = 0
using a = 2, b = 1 and c = 12 in the quadratic equation. This leads to
x = โ^1 ยฑ^
= โ^1
ยฑ i
- Solve the radical equation
3 x + 7 โ 1 = x. Check all proposed solutions.
Answer. Rewrite the equation as
3 x + 7 = x + 1 and then square both sides to obtain 3 x + 7 = x^2 + 2x + 1, or x^2 โ x โ 6 = 0. Therefore, (x โ 3)(x + 2) = 0, and so x = 3 and x = โ2 are proposed solutions. Plugging these back into the original equation, we find that x = 3 works, while x = โ2 does not. Therefore, the solution is x = 3.
- Solve the equation (2x + 1)^23 + (2x + 1)^13 = 6.
Answer. Let u = (2x + 1)^13 to convert this to the equation u^2 + u โ 6 = 0. Therefore, (u + 3)(x โ 2) = 0 and so u = โ3 or u = 2. Therefore, 2x + 1 = โ27 or 2x + 1 = 8. This leads to solutions x = โ14 and x = 7/2.
- Solve the inequality | 3 x โ 6 | > 15.
Answer. The absolute value inequality | 3 xโ 6 | > 15 implies that 3xโ 6 > 15 or 3xโ 6 < โ15. Solving these individually we find that x > 7 or x < โ3.
- Solve the inequality x^ + 2 x โ 5
โค 2 and write the answer in interval notation.
