1 A 1 2 a b h b 2 I Ne R Nr r 3 C 5 9 F 32 F 4 1 p 1 q 2 R q 5 I E R r n n Q

# 1 a 1 2 a b h b 2 i ne r nr r 3 c 5 9 f 32 f 4 1 p 1

• Test Prep
• 5

This preview shows page 4 - 5 out of 5 pages.

1. A = 1 2 ( a + b ) h ; b 2. I = Ne R + Nr ; r 3. C = 5 9 ( F - 32); F 4. 1 p + 1 q = 2 R ; q 5. I = E R + r / n ; n Q. Find the solution set of the following quadratic equations by factoring. 1. x 2 - 3 x - 4 = 0 2. 4 x 2 - 25 x + 6 = 0 3. x 2 - 4 x = 12 4. 3 x 2 + 11 x = 4 5. 12 x 2 - 5 x = 2 6. x 2 = 6 x 7. 3 x 2 = 2 x 8. x + 6 = 12 x 2 9. 12 = 20 x 2 - x 10. 36 x 2 + 7 x = 15 11. 45 x 2 - 30 + 29 x = 0 12. x 2 + 3 ax = 10 a 2 13. 12 x 2 + ax - 6 a 2 = 0 14. b 2 x 2 + 2 bx = 3 abx + 6 a jlescaner summer2014 Math Exercises (for Math 11 and Math 17) Page 5 R. Find the solution set of the following quadratic equations using the quadratic formula. 1. x 2 - 5 x - 24 = 0 2. x 2 + 5 x - 36 = 0 3. x 2 + x = 30 4. 5 x 2 - 17 x + 12 = 0 5. 24 x 2 + x = 10 6. x 2 - 10 x - 2 = 0 7. 7 x 2 + 8 x = 1 8. x 2 - 12 x + 40 = 0 9. x 2 + 29 = 4 x 10. 16 x - 5 x 2 = 15 S. Reduce the following equations to the quadratic form to find the solution set. 1. x 4 + 9 = 10 x 2 2. x 4 - x 2 - 2 = 0 3. x 4 + 14 x 2 - 32 = 0 4. x 6 + 26 x 3 - 27 = 0 5. x 6 + 7 x 3 - 8 = 0 6. 3 x - 2 - 10 x - 1 + 8 = 0 7. ( x 2 - 9) 2 - 8( x 2 - 9) - 128 = 0 8. ( x 2 - x ) 2 - 14( x 2 - x ) + 24 = 0 9. ( x 2 + 2 x ) 2 - 8( x 2 - 2 x ) + 15 = 0 10. (2 x 2 + 7 x ) 2 + 10(2 x 2 + 7 x ) + 24 = 0 11. x x - 3 2 + 3 x x - 3 - 10 = 0 12. x + 1 x - 2 2 = 2 x + 1 x - 2 + 3 13. 3 x - 1 x + 5 2 - 21 = 4 3 x - 1 x + 5 14. x + 3 2 x - 1 + 6 2 x - 1 x + 3 = 5 15. x + 7 x - 2 - 3 = 4 x - 2 x + 7 T. Find the solution sets of the following equations. 1. 2 x + 17 - 4 x - 9 = 0 2. 2 x + 5 - 3 x - 14 = 0 3. x 2 + 7 x - 4 - 10 x + 6 = 0 4. 1 + 2 x + 1 = x 5. 5 + 4 x + 5 = 2 x 6. 2 x 2 - 3 x + 1 = 2 x - 4 7. 3 x + 1 = 2 + 2 x - 6 8. 3 x 2 - 5 x - 8 + 4 = 2 x 9. 4 x + 5 - 2 = 2 x + 3 10. 2 x + 7 + 1 = 3 x + 9 11. 4 x - 3 + 2 x - 2 = 7 x + 4 12. x 2 - x - 2 - x 2 - 2 x - 2 = 1 U. Find the solution sets of the following inequalities using the interval method. 1. x 2 + 7 x + 12 > 0 2. x 2 3 x + 10 3. 4 x 2 + 3 13 x 4. 4 x < 3 x 2 - 15 5. - 6 x 2 x - 12 6. 11 x - 6 3 x 2 7. x - 6 x + 2 > 5 8. x + 9 x + 3 < - 2 9. x 2 - x + 2 x + 1 x - 1 10. 2 x 2 - 1 x + 3 < 2 x - 1 11. x 2 + x + 3 x + 1 3 + x 12. | - 5 x + 4 | > 6 13. x + 1 x - 2 < 2 14. 2 x - 1 2 x + 3 2 15. | 4 x - 3 | < x 16. x + 1 x - 1 x jlescaner summer2014 • • • 