LECTURE 1 : ALGEBRA 01

I. QUADRATIC EQUATION :

1. ME Board Problem Find the value of x in the equation 6x2 + 11x + 3 = 0?

A. -1/3, 3/2 C. 1/3, 3/2B. -1/3, -3/2 D. 3/2, -1/3

Solution:

x = −b±√b2−4 ac

2a = -1/3 & -3/2

Alternate1:MODE EQN 3(QUAD)Enter data6 11 3 Press = -1/3Press = -3/2

Alternate2:RE, substitution using CALCEnter equation: 6x2 + 11x + 3Press CALC X?Enter the item from the choices, example -1/3Press = 0 (solution)

2. ME Board ProblemFind the value of k of the equation x2 + kx + 4 = 0 so that the roots are equal.A. 2 C. 6B. 4 D. 8

Solution:

x=−k ±√k2−4(1)(4 )

2(1)

Equate discriminant to zero

k 2−4 (1)(4)=0 , so k = 4 or -4

Alternate: MODE EQN 3(QUAD)Enter data1 2(letter A) 4

Press = −1+√3 iPress = −1−√3 i (roots are not equal; its not A)

Repeat the same step for Letter B, C, or D.

3. ME Board ProblemThe roots of the quadratic equation are 2 and 3, what is the quadratic equation? A. x2 + 5x + 6 = 0 C. x2 + 5x - 6 = 0B. x2 - 5x + 6 = 0 D. x2 + 7x + 6 = 0

Solution:(x-2)(x-3) = 0x2 - 5x + 6 = 0

Alternate1:MODE EQN 3(QUAD)

For A. 1 5 6 x ≠ 2, 3

For B. 1 -5 6 x = 2, 3(answer)

Alternate2:RE, substitution using CALC (see #1)

II . THEORY OF EQUATIONS :

4. ME Board ProblemIf the polynomial x3 + 4x2 – 3x + 8 is divided by x – 5, what is the remainder?

Solution:Remainder theorem:Given the expression f(x), and an expression (x – r).The remainder R, if f(x) is divided by (x-r) is given by: R = f(r)

R = f(5) = (5)3 + 4(5)2 – 3(5) + 8 = 218 or evaluate using CALC

5. ME Board Problem Find the value of k for which x + 4 is a factor of x3 + 2x2 –

7x + k =0.A. 2 C. 4

B. 3 D. 5

Solution:In order for x + 4 to be a factor of x3 + 2x2 – 7x + k the remainder must be zero.

R = (-4)3 + 2(-4)2 – 7(-4) + k = 0 , so k = 4

Alternate:Using SOLVE function for two or more variablesEnter x3 + 2x2 – 7x + A = 0, A (calcu will solve for A)Press SHIFT SOLVE X?Enter -4Press = Solve for APress = A = 4

III. BINOMIAL THEOREM:

6. ME Board Problem: Find the term involving x9 in the expansion of (x2 + 1/x)12.

Solution:

The (e+1)th term of binomial expansion: nC e An-e Be

12Ce (x2)12-e (x-1)e

12Ce x24-2e x-e

12Ce x24-3e where 24 – 3e must be 9

24 – 3e = 9, so e = 5

Answer: 12C5 x9 = 792x9

7. ME Board Problem Find the 7th term of the expansion (2x – y)11.

Solution:11C6 (2x)5(-y)6 (the exponent of the last term is one

less than its position)8. ME Board Problem

Find the sum of the coefficients of the expansion of (2x + y – z)8.

Solution:(2 + 1 – 1)8 = 256

IV. LOGARITHMIC FUNCTIONS :

9. ME Board Oct. 1997:

Find the value of x if log12x = 2.A. 150 C. 144B. 130 D. 140

Solution:122 = x

Alternate1:

Enter log12 x=2Press SHIFT SOLVE Solve for XPress = X = 144

Alternate2:R.E. using CALC function

Transpose all values in one side log12 x−2

Press CALC X?Enter 150 (letter A)Press = 0.0164 (not zero)Repeat steps for B, C, or D until the result is zeroThe item that gives zero is the answer

10. ME Board Problem The log10 (8)(6) is equal to :

A. log108 + log106 C. log108 – log106B. log108log106 D. log108 / log106

Solution:

log b xy=logb x+ logb y

Alternate:R.E. evaluate the expressions and compare the values

11. ME Board Problem If log 2 = x and log 3 = y, find log 1.2 in terms of x and y.

A. x + y + 1 C. x – y – 1B. 2x + y – 1 D. xy + 1

Solution:Log 1.2 = log (12/10) = log (2x2x3/10)

= log 2 + log 2 + log 3 – log 10= 2log 2 + log 3 – log 10= 2x + y – 1

Alternate:R.E. Evaluate log 1.2 0.079 *

Evaluate option A log 2 + log 3 + 1 1.778Evaluate option B 2log 2 + log 3 – 1 0.079 *Evaluate option C log 2 – log 3 – 1 -1.176Evaluate option D (log 2) (log 3) + 1 1.144

V. PROGRESSION

A. Arithmetic Progression12. ME Board Problem Find the last term of the series of numbers 2, 5, 8, …to the

8th term.

Solution:an = a1 + (n – 1)d

a1 = 2n = 8d = 5 -2 = 3

a8 = 2 + (8 – 1)(3) = 23

Alternate:MODE 3(STAT) 2(A+BX)X Y1 22 5

Press AC

Press 8 SHIFT 1(STAT) 5(REG) 5(ŷ ) = 23

13. ME Board Problem Find the 30th term of an A.P. 4,7,10,…

Solution:an = a1 + (n – 1)d = 91

Alternate:Same alternate as number 12

14. ME Board Problem Find the sum of the odd numbers from 1 to 61.

Solution:

S = ( n2 ) (a1+an )a1 = 1an = 61n = ?

solve for nan = a1 + (n – 1)d61 = 1 + (x – 1)(2)n = 31

so S = 961

alternate:using summationn (can be used only if the first term, and number of terms are known)

S = ∑1

n

an

S = ∑1

n

(a¿¿1+(x−1)d)¿ = S =

∑1

31

(1+(x−1)(2)¿)¿

15. ME Board Problem Find the 7th term of an AP if the 10th term is 32 and the 15th

term is 47.

Solution:a10 = 32a15 = 47

a7 = ?

32 = a1 + (9)d47 = a1 + (14)dSolve for a, and d.manually, or by using MODE EQN 1a1 = 5

d = 3

a7 = 5 + 6(3) = 23

alternate:MODE STAT A+BXX Y10 3215 47

Press ACEnter 7 SHIFT 1(STAT) 5(REG) 5(Y) = 23

16. ME Board Problem How many terms of the progression 3, 5, 7, … must be

taken in order that their sum will be 2600?

Solution:

S = ( n2 ) (a1+an )an = a1 + (n – 1)d = 91

2600 = ( n2 ) (3+(3+(n−1)(2))) solve for

nn = 50

B. Geometric Progression 17. ME Board Problem Find the sum of the first 10 terms of the geometric

progression 2,4,8,16,…

Solution:

S = a1( 1−rn1−r )a1 = 2n = 10r = 4/2 = 2

S = 2046

Alternate:Using summation function( can be used only if the first term, and the number of terms are known or can be approximated)

S = ∑1

n

an

S = ∑1

n

(a¿¿1 rn−1)¿

S = ∑1

10

(2(2x−1)) = 2046

18. ME Board Problem If the 3rd term of GP is 20, and the 6th term is 160, what is

the first term?

Solution:an = a1rn-1

20 = a1r2 equatio 1160 = a1r5 equation 2

By dividing equation 2 by equation 1We get 160/20 = r3, so r = 2

From equation 1:a1 = 5

C. Infinite Geometric Progression

19. ME Board Problem Find the sum of the infinite geometric progression 6, -2,2/3,…

Solution:

S = a1( 11−r )

a1 = 6r = -2/6 = -1/3so, S = 4.5

Alternate:

Using summation function( first term is known, but the number of terms is infinite; use 20 as the upper limit, it will give a sufficiently close result)

S = ∑1

n

an

S = ∑1

n

(a¿¿1 rn−1)¿

S = ∑1

20

(6((−1/3)x−1)) = 4.5

20. ME Board Problem What is the fraction in lowest term equivalent to 0.77777… Solution:

The number can be written as the sum of series of number0.7, 0.07, 0.007,...

a1 = 0.7r = 0.07/0.7 = 0.1

S = a1( 11−r ) 7/9

Alternate:Evaluate the items from A to D and compare to 0.7777...

21. ME Board Problem A rubber ball is made to fall from a height of 50 ft and is

observed to rebound 2/3 of the distance it falls. How far will the ball travel before coming to rest if the ball continues to fall in this manner?

Solution:a1 = 50r = 2/3

S = a1( 11−r ) x 2 – 50 250

Alternate: see number 19 for guideD. Harmonic Progression

22. The 3rd term of a harmonic progression is 15 and the 9th

term is 6. Find the 11th term,

Solution:Illustration: for an arithmetic progression

2, 4, 6, 8, ...The corresponding harmonic progression is:1/2, 1/4, 1/6, 1/8, ..

H.P. A.P.a3 = 15 a3 = 1/15a9 = 6 a9 = 1/6a11 = ? a9 = ?

there is no direct formula to calculate for harmonic, use the AP formulas instead and convert to HP by reciprocating.

Using calcu method: MODE STAT A + BX3 1/159 1/6

Press ACEnter 11 SHIFT 1(STAT) 5(REG) 5(Y) = 1/5

So the ninth term of the AP is 1/5, and the corresponding ninth term of the HP must be 1/(1/5) or 5..

Practice Problems

INSTRUCTION: Choose the best answer.

1. ME Board Problem The log10 (8)(6) is equal to :

A. log108 + log106 * C. log108 – log106B. log108log106 D. log108 / log106

Solution:Same as sample no. 10

2. ME Board Problemln 7.18xy equals

A.1.97xy * C. 0.86 xyB. xy D. 7.18xy

Solution:

By property of logarithm log bax=xlogba

Alternate:R.E. (substitute values to x and y, i.e x=1, y=2)ln 7.181*2 equals = 3.886A.1.97*1*2 * = 3.94 C. 0.86*1*2 = 1.72B. 1*2 = 2 D. 7.18*1*2 = 14.36

3. ME Board Problem 38.5 to the x power = 6.5 to the x – 2 power, solve for x

using logarithms.A. 2.70 C. 2.10B. –2.10 * D. –2.02

4. ME Board Problem What is the value of log to the base 10 of 10003.3 ?A. 9.9 * C. 10.9B. 99.9 D. 9.0

5. What is the value of log2 5 + log3 5 ?A. 7.39 C. 3.97B. 3.79 * D. 9.37

6. ME Board ProblemSolve for x: 6x – 4 = 2x + 6A. -5 C. 10B. 2.5 * D. 5

7. ME Board ProblemWhat are the characteristic and mantissa of the common logarithm of 407.09?A. 2, 0.60969 * C. 4, 0.0709B. 4, 0.60969 D. 2, 0.0709

8. ME Board ProblemWhat is the value of x in the expression x – 1/x = 0 ?A. -1 C. 1B. 1, 1/2 D. 1, -1 *

9. ME Board ProblemSolve for x : x2 + x – 12 = 0A. 6, -2 C. 3, -4 *B. 1, 12 D. 4, -3

10. ME Board ProblemSolve for x : Ax – B = Cx + D

A. * C.

B. D. 11. ME Board Problem If 4x3 + 18x2 + 8x – 4 is divided by 2x + 3, compute the

remainder.A. 11* C. -8B. -9 D. 10

12. The sum of roots of the equation x2 – 3x – 10 = 0 is A. 3 * C. -10 B. -3 D. 10

13. The logarithm of the quotient M/N and the logarithm of the product MN is equal to 1.55630251 and 0.352182518 respectively. Find the value of M.A. 6 C. 8B. 7 D. 9 *

14. What is the value of log6845? A. 5.84 C. 3.76 *

B. 2.98 D. 4.48

15. Find the numerical coefficient of the term involving b4 in the expansion of (a2 – 2b)10.

A. 3360 * C. 6330

B. 3630 D. 6033

16. Given f(x) = (x – 4)(x + 3) + 4, when f(x) is divided by (x – k), the remainder is k. Find k. A. -12 C. 4 *B. 8 D. 2

17. What is the value of x if log 2 to the base 2 plus log x to the base 2 is equal to 2?A. 4 C. -2B. 2 * D. -1

18. The equation whose roots are the reciprocals of the roots of the equation, 2x2 – 3x – 5 = 0

A. 2x2 – 5x – 3 = 0 C. 5x2 + 3x – 2 = 0 * B. 5x2 – 2x – 3 = 0 D. 3x2 – 5x – 2 = 0

19. Solve for x in the equation: 3x + 9x = 27x.A. 0.438 * C. 0.416B. 0.460 D. 0.482

20. Find the value of k in the quadratic equation 3x2 – kx + x – 7k = 0 if 3 is one of the roots.A. 3 * C. 5B. 9 D. 14

21. What is the sum of the coefficients of the expansion of (2x – 1)20 A. 0 * C. 2B. 1 D. 3

22. What is the 5th term of the expansion of (ex + e-x)8 ?A. 60 C. 70 *B. e4x D. sinh x

23. ME Board Problem Find the value of x which will satisfy the following

expression A. 3/2 C. 9/4

B. 9/2 D. no solution *

24. ME Board Problem

Solve for x if A. 4, -5 * C. -4, 5B. -4, -5 D. no solution

25. If log 9! = 5.5598, what is the value of log 10!?A. 5.6598 C. 6.5598 *B. 9.6855 D. 8.6955

26. What is the middle term in the expansion of (x2 – 5)8?A. 47350x8 C. 45370x8

B. 43750x8 * D. 47530x8

27. ME Board ProblemFactor the expression x2 + 6x + 8 as completely as possible.A. (x + 4)(x + 2) * C. (x - 4)(x - 2) B. (x - 4)(x + 2) D. (x + 4)(x - 2)

28. Which of the following is a factor of 3x3 + 2x2 – 32. A. x – 2 * C. x + 2

B. x – 3 D. x + 3

39. If the roots of the equation are -1, 2 and 4, what is the equation?A. x3 – 5x2 + 2x + 8 = 0 * C. x3 – 5x2 - 3x + 6 = 0B. x3 – 4x2 + 3x + 8 = 0 D. x3 – 4x2 + 2x + 6 = 0

30. Find the value of k so that x + 2 is a factor of 3x3 – kx2 + 2x – 4.A. 8 C. -8 *B. 28 D. -28

31. Form a quartic equation with rational coefficients having 3

- and 1 + 2 as roots.A. x4 + 8x3 + 12x2 – 28x – 49=0 C. x4 - 8x3 - 12x2 – 28x – 49 = 0B. x4 - 8x3 + 12x2 + 28x – 49= 0*D. x4 + 8x3 +12x2 +28x + 49 = 0

32. The expression x4 + ax3 + 5x2 – bx + 6 when divided by x – 2 leaves the remainder 16, and when divided by x + 1 leaves the remainder 10. Find the values of a and b.A. 5, 7 C. -5, -7

B. 5, -7 D. -5, 7 *

33. Two engineering students attempt to solve a problem that reduces to a quadratic equation. One of the students made a mistake only in the constant term of the quadratic equation and gives an answer of 8 and 2 for the roots. The other student solving the same problem made an error in the coefficient of the first degree term only and gives his answer as -9 and -1 for the roots. If you are to check their solutions, what would be the correct quadratic equation?

A. x2 + 10x + 9 = 0 C. x2 – 10x – 9 = 0 B. x2 + 10x – 9 = 0 D. X2 – 10x + 9 = 0 *

34. Find the LCM of the numbers 15, 21, 36.A. 1260 * C. 3B. 9 D. 36

35. What is the GCF of 27, 81, and 108?A. 27 * C. 81B. 324 D. 3

36. In the expansion of ( x + 4y )12 , the numerical coefficient of the 5th term is :A. 63,360 C. 506,880

B.126,720 * D. 253,440

37. What is the sum of the coefficients of the expansion of ( 2x – 1 )20 .A. 0* C. 2B. 1 D. 3

38. Determine the sum of the positive valued solution to the simultaneous equations : xy = 15 , yz = 35 , zx = 21.A.15 C. 17B.13 D. 19

39. ME Board ProblemSolve for x : 10x2 + 10x + 1 = 0A. – 0.113 , – 0.887 * C. – 0.331 , – 0.788B. – 0.113 , – 0.788 D. – 0.311 , – 0.887

40. ME Board ProblemSolve for x : x-6/8 = 0.001A. 10 C. 100B. 0 D. 10,000 *

41. If 3a = 7b, then 3a2 / 7b2 is equal toA. 1 C. 7/3 *B. 3/7 D. 49/9

42. If 1/x = a + b, 1/y = a – b, then x – y is equal to A. 2b/(b2 – a2) * C. 2a/(a2 – b2)B. 1/2a D. 1/2b

43. If f(x) = 10x + 1, then f(x + 1) – f(x) is equal to A. 1 C. 11B. 9(10x) * D. 10x + 1

44. If 10x = 4, then the value of 102x + 1 is A. 26 C. 40B. 160 * D. 900

45. If x + y = m and xy = n, then the value of (x – y)2 is A. m2 – 4n * C. m2 – 2nB. m2 + 2n D. m2 + n2

46. -3x2 – 3x + 1 = -3(x + a)2 + b is an identity in x. What is the value of a?A. 1/2 * C. 7/4B. – 1/2 D. -7/4

47. Find the remainder when x4 – 10x2 – 9x – 20 is divided

by x – 4 .

A. 40 C. 50 B. 30 D. 20

48. One factor of a3 - 2a2 – a + 2 is a =1, find the other factors.A. ( a – 2 ) ( a + 1 ) * C. ( a + 2 ) ( a – 1 )B. ( a + 2 ) ( a + 1 ) D. ( a – 2 ) ( a – 1 )

49. Find the value of k so that the remainder upon dividing

by is zero.A. 4 C. -48 *

B. 48 D. 64

50. What is the remainder when 8100 is divided by 5?A. 1 * C. 3B. 2 D. 4