1.SOLVE WITHOUT TABLE: log27 1/2 + log 8 1/2 - log 125 1/2
DIVIDE BY
log 6 - log 5 = 3/2

2. IF ALPHA AND BETA ARE D ROOTS OF THE EQUATION, X SQUARE + 5X-2=0

THEN FIND THE VALUE OF (I)ALPHA SQUARE+ BETA SQUARE (II) 1/BETA + 1/ALPHA


3 FROM DIFFERENTIATIONSOLVE Y = (X SQURE + 4) 9X RAISE TO POWER 3 + 4)

bilms:

1.SOLVE WITHOUT TABLE: log27 1/2 + log 8 1/2 - log 125 1/2
DIVIDE BY
log 6 - log 5 = 3/2

2. IF ALPHA AND BETA ARE D ROOTS OF THE EQUATION, X SQUARE + 5X-2=0

THEN FIND THE VALUE OF (I)ALPHA SQUARE+ BETA SQUARE (II) 1/BETA + 1/ALPHA


3 FROM DIFFERENTIATIONSOLVE Y = (X SQURE + 4) 9X RAISE TO POWER 3 + 4)

Eqn 2 looks incorrect, x2 +5x-2=0

ITS CORRECT,

bilms:

1.SOLVE WITHOUT TABLE: log27 1/2 + log 8 1/2 - log 125 1/2
DIVIDE BY
log 6 - log 5

= 3/2

Soln

log27 1/2 + log 8 1/2 - log 125 1/2/
log6 – log5

log3 3/2 + log2 3/2 – log 5 3/2/
log 6- log5

3/2 log 3 + 3/2 log 2 – 3/2 log 5/
Log 6 – log5

3/2 { log3 + log2 – log5}/
Log6 – log5

3/2 ( log 6 – log5)/
Log6 – log5

= 3/2


2. IF ALPHA AND BETA ARE D ROOTS OF THE EQUATION, X SQUARE + 5X-2=0

THEN FIND THE VALUE OF (I)ALPHA SQUARE+ BETA SQUARE (II) 1/BETA + 1/ALPHA
[quote][/quote]

Soln

X*2 +5x – 2 = 0

X*2 + 5x = 2

Add the square of half the quoficient of x to both sides;
This is what you get;

X*2 + [5]*2 + 5x = 2 + (5)*2
(2) (2)


(X + 5)*2 = 2 + 25
2 4

Am I on course?
Busy now to finish it up soon.

Catch ya.

i guess i will continue where you left off infogenius lol

from (x+5/2)squared = 33/4
(x+5/2) = +/- sqrt(33/4)
x1 = alpha = + [sqrt(33) - 5] / 2
x2 = beta = - [sqrt(33) - 5]/ 2

I) alpha squared + beta squared = plug and chug
II) 1/alpha + 1/beta = plug and chug

I am lurking.

Let me wrap up,

Let Alpha = & and beta = $
Therefore from the equation
X*2 +5x – 2 = 0
We will have a= 1, b = 5, c = -2
Sum of roots, & + $ = b = 5 = 5
a 1

Product of roots &$ = c = -2 = -2
a 1

1. (&*2 + $*2) =[ No factors or maybe it is supposed to be
( & + $ )*2].
If it is (& + $)*2 then (& + $) = 5
Then (& + $)*2 = 5*2 = 25.

2. 1 + 1 = & + $ = 5 = -2.5
& $ &$ -2


3. FROM DIFFERENTIATIONSOLVE Y = (X SQURE + 4) 9X RAISE TO POWER 3 + 4)
Soln

Y = (x*2 + 4) (9x*2 + 4) if I got you right
Let’s go……….
Let (x*2 + 4) = U
And (9x*2 + 4) = V
Therefore Y = UV

If (x*2 + 4) = U ;then du = 2x
dx
And (9x*2 + 4) = V; then dv = 18x
dx

Product rule = dy = Udv / + Vdu
dx dx dx


Putting parameters we have
dy = (x*2 + 4)18x + (9x*2 + 4)2x
dx

Expanding dy = 18x*3 +56x +18x*3 + 8x
dx

dy = 36x*3 + 64x
dx

dy = 4x [9x*2 + 16]
dx

Let me see if I can simplify further.
Ditto!