QUADRATICS - I1

 

ALGEBRA - I1

 If │4x – 7 │= 19; Find  x

 

Solution

 

±(4x – 11 )= 19

 

4x -11= 19   OR   –(4x-11) = 19

 

4x – 11 = 19    OR   -4x+11=19

 

4x = 19 + 11    OR   -4x= 19-11

 

4x = 30     OR   -4x  =   8

4      4               -4        -4

 

x = ⁷/2  OR  x = -2

 

 

Hence x = ⁷/2  OR  x = -2

ARITHMETIC - I1

 Evaluate 18 x 369 + 331 x 18.

 

Solution

 

= 18 x 369 + 331 x 18.

 

= 18 x (369 + 331)     factoring out the common number

 

= 18 x 700   

 

= 12600      [after multiplying 18and 7 and adding two zeros on the answer]

 

Hence 18 x 369 + 331 x 18= 12600

 

TRY THIS………….

 

 

Evaluate 137 x 516 + 484 x 137.

ALGEBRA - I1

 

The sum of four consecutive numbers is 474. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+1	n+2	n+3	474

 

Then, n + (n+1) + (n+2) + (n+3) = 474

 

4n + 1+2+3= 474

 

4n + 6 = 474

 

4n = 474 – 6

 

4n = 468 

 

4n = 468

 4        4

 

n = 117

 

3rd number = n +2

 

                   = 117 + 2

 

                   = 119

 

Hence the 3rd number is 119

 

TRY THIS………………….. 

 

The sum of four consecutive numbers is 1006. Find the largest number.

LOGARITHMS - I1

 Evaluate Log₂(512 ÷ 8).

 

Solution

 

= Log₂(512 ÷ 8)

 

= Log₂512 -  Log₂8          (applying the product rule)

 

= Log₂2⁹ -  Log₂2³             ( 512= 2⁹ and 8=2³ )

 

= 9Log₂2 -  3Log₂2       ( remember  Log_aa^c = cLog_aa )

 

= (9 x 1) -  (3 x 1)          ( remember  Log_aa = 1 )

 

= 9 - 3

 

= 6 answer

 

 

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

 

Evaluate Log₂(2048 - 32).

 

FUNCTIONS -I1

 If f(x) = |2x - 10| evaluate f(-70)

 

Solution

 

f(x) = |2x - 10|

 

f(-70) = |(2 x -70) - 10|

 

f(-70) = |-140 - 10|

 

f(-70) = |-150| = 150 (since any number out of absolute signs is +ve).

 

Hence f(-70) = 150

 

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

 

If f(x) = |5x - 300| evaluate f(-60)

LOGARITHMS G1

 

Evaluate Log₂(512 ÷ 8).

 

Solution

 

= Log₂(512 ÷ 8)

 

= Log₂512 -  Log₂8          (applying the product rule)

 

= Log₂2⁹ -  Log₂2³             ( 512= 2⁹ and 8=2³ )

 

= 9Log₂2 -  3Log₂2       ( remember  Log_aa^c = cLog_aa )

 

= (9 x 1) -  (3 x 1)          ( remember  Log_aa = 1 )

 

= 9 - 3

 

= 6 answer

 

 

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

 

Evaluate Log₂(2048 - 32).

ALGEBRA G1

 

Given that one of the roots of the equation 3x² + m(x+1) + 5 = 0 is 2, find m.

Solution

Substitute x=2, in the above equation.

3(2)² + m(2+1) + 5 = 0

(3x4) + (m x 3) + 5 = 0

12 + 3m + 5 = 0

3m + 17 = 0

3m = -17

3m = -17

3          3

 

m=^-17/3

Hence m=^-17/8

TRY THIS…………………………..

Given that one of the roots of the equation 4x² + m(x+1) + 10 = 0 is 8, find m.

GEOM. PROGRESSION G1

 The 1st term of a geometric progression is 7 and the 5th term is 112. Find i)the common ratio. ii) 10^th term

 

Solution

 

G₁=7, G₅=112, n=5

 

i) G_n = G₁r^n-1;

 

G₅ = G₁r^5-1;

 

G₅ = G₁r⁴;

 

112 = 7 x r⁴;

 

112 = 7r⁴;

 7       7

 

16 = r⁴;  Finding the fourth root of 16,

 

r = 2

 

 

 

Hence the common ratio is 2.

 

ii)  10^th term

 

G_n = G₁r^n-1;

 

G₁₀ = G₁r^10-1;

 

G₁₀ = G₁r⁹;

 

G₁₀ = 7 x 2⁹;

 

G₁₀ = 7 x 512;

 

G₁₀ = 3584

 

Hence the eighth term is 3584

 

TRY THIS………………………….. 

 

The 1st term of a geometric progression is 2 and the 6th term is 486. Find i) the common ratio. ii) 11^th term.

 

EXPONENTIALS H2