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)