FACTORIZATION C2

Factorize 2x² - 7x - 30

Solution

= 2x² - 7x – 30.  We split the middle term (-7x) to be (-12x + 5x).

= 2x² - 12x + 5x – 30     (-7x =-12x + 5x).

= (2x² -12x) + (5x - 30)

= 2x(x - 6) + 5(x - 6)

= (2x + 5) (x -6)

Hence2x² - 7x – 30 = (2x + 5) (x -6) answer.

TRY THIS…..

Factorize 4x² - 3x – 10.

DIFFERENCE OF 2 SQUARES C1

Evaluate 780² – 680²

Solution

780² – 680² = (780 + 680)( 780 – 680)

                   = (1460)( 100)

                   = 146000

Hence 780² – 680² = 146000

LOGARITHMS B38

Evaluate the following giving your answer in standard form.

38.3 x 10^-4

   4 x 10^-20

Solution

=  38.3 x 10^-4

         4 x 10^-20

=  38.3  x   10^-4

     4         10^-20

=  9.575 x 10^{-4 –(-20)}

=  9.575 x 10^{-4 + 20}    

=  9.575 x 10¹⁶     

Hence   38.3 x 10^-4   = 9.575 x 10¹⁶     

                 4 x 10^-20

RADICALS C10

INEQUALITIES C1

2x – 3 ≤ x + 15 ≤ 6x

Solution

2x – 3 ≤ x + 15 and  x + 15 ≤ 6x

2x – x ≤ 3+ 15 and  15 ≤ 6x - x

x ≤ 18 and  15 ≤ 5x (divide by5 both sides)

x ≤ 18 and  3 ≤ x

x ≤ 18 and  x ≥ 3  answer

RADICALS C9

FACTORIZATION C11

Factorize 2x² - 15x + 7.

Solution

= 2x² - 15x + 7. We split the middle term (-15x) to be (-x - 14x).

= 2x² -x - 14x + 7

= (2x² – x) – (14x + 7)

= x(2x – 1) – 7(2x - 1)

= (x - 7) (2x - 1)

Hence 2x² - 15x + 7= (x - 7) (2x - 1)answer.

TRY THIS…..

9x² - 8x - 1.

MATRIX B1

PROBABILITY C14

Given the interval of 1-10 inclusive, Find the probability of having an even number or a prime number.

Solution

This is a non-mutually exclusive event.

n(S) = number of sample space = 10

Let P(E)= probability of even numbers = ⁵/10

Let P(R)= probability of prime  numbers = ⁴/10

P(EnR) = probability of having both even and prime numbers= ¹/10

∴ P(EuR) = P(E) + P(R)  - P(EnR).

 P(EuR) =  5  +  4 – 1

                10    10   10

P(EuR) =  9  -   1 

                10    10  

P(EuR) =  8 

                10       

Hence probability of an even number or a prime number is ^8/10.

TRY THIS..............

Given the interval of 11-20 inclusive, Find the probability of having an odd number or a prime number.

CIRCLES C8

In the figure below, find the value of <FGE.

Solution

<FEG = <HFG

∴ <FEG = 54⁰ [angles in alternate segments are equal.]

Consider ∆GEF

<EFG + <FGE+ <GEF = 180⁰ [total degrees of a triangle]

 (39⁰ + 54⁰) + <FGE + 54= 180⁰

93⁰ + <FGE + 54⁰ = 180⁰

147⁰ + <FGE= 180⁰

<FGE= 180⁰ - 147⁰

<FGE= 33⁰  

∴ <FGE = 33⁰  

LOGARITHMS B37

If log_x625 – log₂64 = -4; find x

solution

log_x625 – log₂64 = -2

log_x625 – log₂2⁶ = -2

log_x625 – 6log₂2 = -2

log_x625 – 6 x 1 = -2

log_x625 – 6 = -2

log_x625 = -2 + 6

log_x625 = 4

625 = x⁴

5⁴ = x⁴      Powers cancel out

x = 5

Hence x = 5.

FACTORIZATION C10

Factorize 3x² - 16x + 5.

Solution

= 3x² - 16x + 5. We split the middle term (-16x) to be (-x - 15x).

= 3x² - x - 15x + 5.

= (3x² – x) – (15x + 5)

= x(3x – 1) – 5(3x - 1)

= (x - 5) (3x - 1)

Hence 3x² - 16x + 5= (x - 5) (3x - 1)answer.

TRY THIS…..

6x² - 17x - 14.

RADICALS C8

EXPONENTS A11

simplify 5

              8^-2/3

Solution

=  5

    8^-2/3

=  5 x 1

           8^-2/3

=  5 x 8^2/3        [Since 1/a^-n = aⁿ]

=  5 x (8^1/3)²        [Since 8^1/3 = cube root of 8 = 2]

=  5 x (2)²        

=  5 x 4

= 20        

          

Hence   5        = 20

              8^-2/3

EXPONENTS A10

Simplify (D¹¹)¹⁰

solution

= (D¹¹)¹⁰

=  D^11x10

=  D¹¹⁰

TRY THIS

Simplify  (G²)¹²

FACTORIZATION C9

Factorize 2x² - 3x - 35.

Solution

= 2x² - 3x - 35. We split the middle term (-3x) to be (+7x - 10x).

= 2x² + 7x - 10x - 35.

= (2x² + 7x) – (10x – 35).

= x(2x + 7) – 5(2x + 7).

= (x - 5) (2x + 7)

Hence 2x² - 3x – 35 = (x - 5) (2x + 7) answer.

TRY THIS…..

8x² - 22x + 5.

RADICALS C7

LOGARITHMS B36

If log_x49 – log₂128 = -5; find x

solution

log_x49 – log₂128 = -5

log_x49 – log₂2⁷ = -5

log_x49 – 7log₂2 = -5

log_x49 – 7 x 1 = -5

log_x49 – 7 = -5

log_x49 = -5 + 7

log_x49 = 2

49 = x²

7² = x²      Powers cancel out

x = 7

Hence x = 7.

CIRCLES C7

In the figure below, find the value of <RUS.

Solution

<RSU = <TRU

∴ <RSU = 54⁰ [angles in alternate segments are equal.]

Consider ∆SRU

<SRU + <RUS+ <RSU = 180⁰ [total degrees of a triangle]

 (46⁰ + 54⁰) + <RUS + 54= 180⁰

100⁰ + <RUS + 54⁰ = 180⁰

154⁰ + <RUS= 180⁰

<RUS= 180⁰ - 154⁰

<RUS= 26⁰  

∴ <RUS = 26⁰  

PROBABILITY C13

A bag contains 6 red stones and 8 blue stones. Two balls are taken from the bag. What is the probability that they are both red?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(R) = n(R)

            n(S)

1st pick = 6/14

2nd pick = 6/14 as well.

P(R) =  6      x    6

          14          14

P(R) =    36    =      4 

             196           49  

Therefore Probability of drawing a red stone is 4/49     

LOGARITHMS B35

If log_x27 + log₂256 = 11; find x

solution

log_x27 + log₂256 = 11

log_x27 + log₂2⁸ = 11

log_x27 + 8log₂2 = 11

log_x27 + 8 x 1 = 11

log_x27 + 8 = 11

log_x27 = 11- 8

log_x27 = 3

27 = x³

3³ = x³      Powers cancel out

x = 3

Hence x = 3.

CIRCLES C6

In the following figure, prove that the angles in the same segment of a circle are equal.

Solution

Given : A circle with centre O and the angles ∠USV and ∠UTV in the same segment formed by the chord UV (or arc UAV)

Required to prove: ∠USV = ∠UTV

Construction : Join OU and OV.

Proof :

Try to remember that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle, hence we have,

∠UOV = 2 ∠USV ...............(i)

and,

∠UOV = 2∠UTV …………...(ii)

Since (i) = (ii), we get,

2∠USV = 2∠UTV

2∠USV = 2∠UTV                 (dividing by 2 both sides)

2             2

∠USV = ∠UTV

∴ ∠USV =∠UTV   proved!

PROBABILITY C12

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of hearts?

Solution

(This is a problem with replacement)

n(E) = 13, n(S) = 52

P(E) = n(E)

           n(S)

1st pick = 13/52

2nd pick = 13/52 as well.

P(2 hearts) =   13      x    13

                       52            52

P(2 hearts) =    169    =      1 

                        2704         16  

Therefore Probability of drawing 2 heart cards is 1/16     

TRY THIS............

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of spades?

LOGARITHMS B34

If log_x625 + log₃81 = 8; find x

solution

log_x625 + log₃81 = 8

log_x625 + log₃3⁴ = 8

log_x625 + 4log₃3 = 8

log_x625 + 4 x 1 = 8

log_x625 + 4 = 8

log_x625 = 8- 4

log_x625 = 4

625 = x⁴

5⁴ = x⁴      Powers cancel out

x = 5

Hence x = 5.

CIRCLES C5

In the figure below, find the value of <HLJ.

Solution

<HJL = <LHK

∴ <HJL = 54⁰ [angles in alternate segments are equal.]

Consider ∆HJL

<JHL + <HLJ+ <HJL = 180⁰ [total degrees of a triangle]

 (44⁰ + 54⁰) + <HLJ + 54= 180⁰

98⁰ + <HLJ + 54⁰ = 180⁰

152⁰ + <HLJ= 180⁰

<HLJ= 180⁰ - 152⁰

<HLJ= 28⁰  

∴ <HLJ= 28⁰  

PROBABILITY C11

A bag contains 6 red balls and 8 blue balls. Two balls are taken from the bag. What is the probability that they are both blue?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(B) = n(B)

            n(S)

1st pick = 8/14

2nd pick = 8/14 as well.

P(B) =  6      x    6

          14          14

P(B) =    36    =      4 

             196           49  

Therefore Probability of drawing a blue ball is 4/49     

TRY THIS ..............

A bag contains 10 purple stones and 12 blue stones. Two stones are taken from the bag. What is the probability that they are both blue?

LOGARITHMS B33

If log_x625 + log₃27 = 7; find x

solution

log_x625 + log₃27 = 7

log_x625 + log₃3³ = 7

log_x625 + 3log₃3 = 7

log_x625 + 3 x 1 = 7

log_x625 + 3 = 7

log_x625 = 7- 3

log_x625 = 4

625 = x⁴

5⁴ = x⁴      Powers cancel out

x = 5

Hence x = 5.

CIRCLES C4

In the figure below, find the value of <MNQ.

Solution

<MQN = <PMN.

∴ <MQN = 52⁰ [angles in alternate segments are equal.]

Consider ∆MQN

<QMN + <MNQ+ <MQN = 180⁰ [total degrees of a triangle]

 (38⁰ + 52⁰) + <MNQ + 52= 180⁰

90⁰ + <MNQ + 52⁰ = 180⁰

142⁰ + <MNQ= 180⁰

<MNQ= 180⁰ -142⁰

<MNQ= 38⁰  

∴ <MNQ= 38⁰