Calculator technique session 1
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The document discusses techniques for solving quadratic equations using a calculator. It provides examples of finding the roots of quadratic equations by inputting the values of a, b, and c into the calculator. It also discusses calculator modes, functions, and setup, including how to use the shift function to solve equations and the remainder theorem to find the remainder when dividing a polynomial by a linear factor.
Calculator technique session 1
- 1. Calculator Techniques Prepared By: Engr. Jamille G. Quindara
- 2. Lesson 1. Calcu Tech/ Theory / Practice
- 3. Quadratic Equation A quadratic equation is a second order polynomial equation in a single variable. Because it is second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions.
- 5. Problem 1. Find the roots of X2 + 5x + 6 a. -2 , -3 b. 10, 5 c. 6, -3 d. -8, 1
- 6. Solution : Mode (5) (3)
- 7. Solution : Mode (5) (3) *Input value of a b and c 1 (=) 5 (=) 6 (=)
- 8. Solution : Mode (5) (3) *Input value of a b and c 1 (=) 5 (=) 6 (=) Ans. -2 , -3
- 9. Problem 2. Find the sum of the roots of the equation 3X2-11x-4=O a. -1/4 b. 10/6 c. 1/4 d. 11/3
- 10. Solution : Mode (5) (3)
- 11. Solution : Mode (5) (3) *Input values of a b and c 3 (=) -11 (=) -4 (=)
- 12. Solution : Mode (5) (3) *Input values of a b and c 3 (=) -11 (=) -4 (=) X1 = 4 , X2 = -1/3
- 13. Solution : Mode (5) (3) *Input values of a b and c 3 (=) -11 (=) -4 (=) X1 = 4 , X2 = -1/3 4 + (-1/3) = 11/3
- 14. Problem 3. The equation whose roots are the reciprocals of the roots of the equation, 2X2 - 3x – 5 = 0 a. 2X2 – 5x – 3 = 0 b. 5X2 – 2x – 3 = 0 c. 5X2 + 3x – 2 = 0 d. 3X2 – 5x – 2 = 0
- 15. Lesson 2: Calculator Functions/ Mode/ Setup
- 16. Default Calculator Mode: Press : ON Shift Mode Setup 1 (Mth I0) 1 (Math0) Mode Setup 1 (Comp)
- 17. Clearing Cache Press : ON Shift 9 3 (All) = (yes) AC
- 18. Press Mode 1: Comp 2: Cmplx 3: Stat 4: Base-N 5: Eqn 6: Matrix 7: Table 8: Vector
- 19. Mode 1. Comp Includes basic computation/basic functions
- 20. Lesson 3. Mode 1 ( Comp)
- 21. LESSON 3.1 SHIFT SOLVE Problem. 1 Solve x in the equation 2log 4 x - log 4 9 = 2 a. 12 b. 10 c. 11 d. 12
- 22. Solution: Input 2log 4 x - log 4 9 = 2
- 23. Solution: Input 2log 4 x - log 4 9 = 2 Alpha XAlpha ) log Alpha Calc
- 24. Solution: Input 2log 4 x - log 4 9 = 2 Press Shift Calc Input Any number from the choices Alpha ) log Alpha Calc
- 25. 225 = Find n log x2 – log 5x = log 20 ;Find x ½(x^2+3x) – ½(x^2-3x) = 21 Find x 4(72n+1) – 10(52n-1) 2(42n) 661/16
- 26. LESSON 3.2 REMAINDER THEORM Problem 2. When you divide X2 – 2x + 2 = 0 by (x – 2), find the remainder. a. 2 b. 4 c. 5 d. 8
- 27. Solution: Input the equation X2 – 2x + 2 Replace X by 2
- 28. Solution: Input the equation X2 – 2x + 2 (2)2 – 2(2) + 2 Replace X by 2
- 29. Solution: Input the equation X2 – 2x + 2 (2)2 – 2(2) + 2 = 2 Replace X by 2
- 30. If (x+3) is a factor of X3 + 3X2 + 4X + K , f ind K a. 11 b. 12 c. 13 d. 14