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Solve Maths by bilms(m): 12:39pm On Aug 07, 2009 |
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) |
Re: Solve Maths by debest1(m): 4:48pm On Aug 07, 2009 |
bilms:Eqn 2 looks incorrect, x2 +5x-2=0 |
Re: Solve Maths by bilms(m): 7:39pm On Aug 07, 2009 |
ITS CORRECT, |
Re: Solve Maths by infogenius(m): 8:26pm On Aug 11, 2009 |
bilms: 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 = 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. |
Re: Solve Maths by C2H5OH(f): 8:52pm On Aug 11, 2009 |
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 |
Re: Solve Maths by infogenius(m): 9:56am On Aug 12, 2009 |
I am lurking. |
Re: Solve Maths by infogenius(m): 8:36pm On Aug 13, 2009 |
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! |
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