@doubledx & richiez. Keep d gud work going jor

nice.one:
wow! i really missed out on this maths frenzy maths is so fun that it became a past time back then in school. right now am just wallowing in nostalgia of the abstract/complex maths i did during my MSc in control systems engineering.i would love to be part of this but right now am way 2busy. i hail all the guru that is making things happen. maybe i will be recommending you guys for the Melina maths prize .
just hv these few contribution to make
for;
i) X^2=2^X
apart from numerical solution (graphical, etc) it can also be solved by differentiating twice and solving for X

ii) 3^X +2^X=5^x this can also be solved by playing around diff.

ans x=4 or x=2

solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Bless

Bolaji 16: solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Bless

the question is not clearly interpretable pls retype in a way that we can understand so we can help u out quickly

Bolaji 16: solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Bless

Is this what you meant √(x ) - √(x - 1) = 1? If it is then here is the solution>

√(x ) - √(x - 1) = 1
rearranging> √(x - 1) = √(x) - 1
Squaring both sides to get rid of the roots>
[√(x - 1)]^2 = (√(x) - 1)^2
x - 1 = x - 2√(x) + 1
-1 -1 = - 2√(x)
-2 = -2√(x)
1 = √(x)
Squaring both sides again yields
[√(x)]^2 = 1^2
x = 1

^Nice one

yungryce: @doubledx & richiez. Keep d gud work going jor

Thanks bruv, I have solved the remaining questions. I'll take out time to type and post them later 2day or 2mrrw. 1luv

Ghettoguru:

Is this what you meant √(x ) - √(x - 1) = 1? If it is then here is the solution>

√(x ) - √(x - 1) = 1
rearranging> √(x - 1) = √(x) - 1
Squaring both sides to get rid of the roots>
[√(x - 1)]^2 = (√(x) - 1)^2
x - 1 = x - 2√(x) + 1
-1 -1 = - 2√(x)
-2 = -2√(x)
1 = √(x)
Squaring both sides again yields
[√(x)]^2 = 1^2
x = 1

thank u very much bro..

Help me check dis out

^Solution

2√(x + 4) - x = 1
2√(x + 4) = x + 1
Square both sides
[2√(x + 4)]^2 = (x + 1)^2
4(x + 4) = x^2 + 2x + 1
4x + 16 = x^2 + 2x + 1
Rearrange and collect like terms =>

x^2 - 2x - 15 = 0
factorizing =>
x^2 - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5) = 0
(x - 5)(x + 3) = 0
:. x - 5 = 0 or x + 3 = 0
x = 5 or -3

Great job Ghetoguru and double dy/dx

doubleDx: ^Solution

2√(x + 4) - x = 1
2√(x + 4) = x + 1
Square both sides
[2√(x + 4)]^2 = (x + 1)^2
4(x + 4) = x^2 + 2x + 1
4x + 16 = x^2 + 2x + 1
Rearrange and collect like terms =>

x^2 - 2x - 15 = 0
factorizing =>
x^2 - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5) = 0
(x - 5)(x + 3) = 0
:. x - 5 = 0 or x + 3 = 0
x = 5 or -3

Sorry my esteemed gurus but doing a check with -3 does not solve the equation
Please check

biolabee:

Sorry my esteemed gurus but doing a check with -3 does not solve the equation
Please check

No probs bruv, check this out =>

2√(x + 4) - x = 1
2√(x + 4) = x + 1

For x = -3 on the LHS

2√(x + 4)
= 2 . [±√(-3 + 4)]
= 2. [±√(1)]
= 2 .(+ 1 ) or 2. (- 1)
= 2 or -2

For x = -3 on the RHS

= x + 1
= -3 + 1
= - 2

It's true for x = -3 that LHS = RHS with a value of (-2)
Note=> (.) is for multiplication!

nice one

we learn everyday

doubleDx:

No probs bruv, check this out =>

2√(x + 4) - x = 1
2√(x + 4) = x + 1

For x = -3 on the LHS

2√(x + 4)
= 2 . [±√(-3 + 4)]
= 2. [±√(1)]
= 2 .(+ 1 ) or 2. (- 1)
= 2 or -2

For x = -3 on the RHS

= x + 1
= -3 + 1
= - 2

It's true for x = -3 that the LHS = RHS with a value of (-2)
Note=> (.) is for multiplication!

..u are nt solving my questions ni?...kindly asist me na

Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks

Sorry bruv, we miss this one =>

Solution to Question 1a

(1 - x)^(-1/2)

Since =>
(1 + r)^k = 1 + kr + k(k - 1)(r)^2/2! + k(k - 1)(k - 2)(r)^3/3! + ....

r = (-x) and k = -1/2 in this case. Substituting yields =>

= 1 + (-1/2)(-x) + (-1/2)( -3/2)(-x)^2/2! + (-1/2)(-3/2)(-5/2)(-x)^3/3! + ....
= 1 + x/2 + 3x^2/8 + 15x^3/8(6) + ...
Hence, the first 4 terms of the expression (1 - x)^(-1/2) =>

= 1 + x/2 + 3x^2/8 + 5x^3/16

I'm a little busy now; will check question 2 later. 1luv man!

Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect!

happy weekend to all my maths gurus......it's been quite a long time....please i would like to ask this:
from a given set of data, is there any mathematical way of calculating the range apart from HIGHEST DIGIT-LOWEST DIGIT........I'm asking because i have never been taught about range.....and whenever i tell my frnds that these is the only way of finding the range from a given data, they will just laugh and say; it doesnt sound like a mathematical method...so plz my mathematicians...help me

ayoope: Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect!

Thanks. Do you have any math problem for us?

Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks

Solution to Question 1b

To prove that √10 = 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]

Evaluating the RHS yields =>
= 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]
= 3 (1.0541)
= 3.162 approx.

√10 can be rewritten as =>

= √[9(1 + 1/9)]
= 3√(1 + 1/9)
= 3 (1 + 1/9)^(1/2)
Expanding the bolded part yields =>
= 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....]
Simplifying further yields =>
= 3[ 1.0541 approx]
= 3.162 approx

Thus,
√10 = 3(1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ...] = 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + .... = 3.162 approx

ayoope: Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect!

thanks ma you are far too kind

@doubleDx, I don't have any.
Been following this thread since the very first day. Just amazing the way you all provide solutions to the questions asked.

Maths is all about accuracy and memory. One you have those two, it will be easy for you to pound on any equation. www.edubabs.com

doubleDx:

Solution to Question 1b

To prove that √10 = 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]

Evaluating the RHS yields =>
= 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]
= 3 (1.0541)
= 3.162 approx.

√10 can be rewritten as =>

= √[9(1 + 1/9)]
= 3√(1 + 1/9)
= 3 (1 + 1/9)^(1/2)
Expanding the bolded part yields =>
= 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....]
Simplifying further yields =>
= 3[ 1.0541 approx]
= 3.162 approx

Thus,
√10 = 3(1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ...] = 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + .... = 3.162 approx

...God bless u man.....tanx alot.....

Fetus: ...God bless u man.....tanx alot.....

Thanks and you're welcome anytime.

ayoope: @doubleDx, I don't have any.
Been following this thread since the very first day. Just amazing the way you all provide solutions to the questions asked.

Thanks a lot ma'am. Your compliment is appreciated!

Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks

Your second question has some issues. I doubt if it's a geometric series! Check this out =>

The geometric sequence is given as => sin2α, -sinαcos2α, sin2αcos^2(2α),.. For Ө<α<π₹/2

The common ratio (r) => a2/a1 or a3/a2 where a1, a2 and a3 are first, second and third term of the series respectively.

Using the 1st and 2nd term,
r = a2/a1 = [-sin(α)cos(2α)]/sin(2α) =>
Since cot(2α) = cos(2α)/sin(2α), then=>
r1 = -cot(2α) sin(α)

Using the 2nd and 3rd term, r = a3/a2
= [sin 2α cos^2(2α)]/[-sin α cos 2α]
Since tan(2α) = sin(2α)/cos(2α), then=>
r2 = -tan (2α)cos^2^(2α)

If the series is geometric, r1 should be equal to r2; but in this case r1≠ r2, thus I doubt if the question is correct! Re-check it bruv.

Sum to infinity for r1 =>
S_∞ = a1/(1 - r1)
If we substitute r1, it yields=>
= sin(2α)/[1 - (-cot(2α) sin(α))]
= sin(2α)/[1 + cot(2α) sin(α)]

or

Sum to infinity for r2=>
S_∞ = a1/(1 - r2)
If we substitute r2, it yields=>
= sin(2α)/[-tan (2α)cos^2^(2α)]
Which is not equal to the result for r1.

Recheck the question. If possible write it out on a paper and scan it. Take care.

Youngsage:
x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

Spoiler Alert: Hello bro, I don't want to cause spoiler-alert but you got your solution wrong. You might be lucky on your answer.

The properties applied on the shaded equation is invalid: log(ab) = log(a) + log(b, which implies that log(a+b) NotEqual log(a)+log(b). I'm glad you've been corrected by other people.

This is a nonlinear equation which can only be solved numerically or by graphical method.

chikis:

Father reduced the
quantity of food
bought for the
family by 10%
when he found
that the cost of
living had increased
by 15%. Thus the
fractional increase
in the family food
bill now is what?

Solve it, maths guru!

Let initial cost of living = C
Let the new increase in cost of living by 15% = 1.15C

He then decreased the quantity of food bought by 10%. In order word, he now purchases 90% of the new cost = 90% * 1.15C = 0.9*1.15C

Thus the fractional increase w.r.t the initial cost (C) = (New Spending - Initial Cost)/(Initial Cost) = (0.9*1.15C - C)/C = (0.9*1.15 - 1)= 0.035

Thus, Fractional Increase = 0.035 or 3.5%

An illiterate, I believe, can follow my follow solutions.

Ehinafe: Richiez:

nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

Richiez, how good are you in solving Dynamic Programming mathematical algorithm? I need to know because I have one here that I am not sure you can get the answer right.

Spoiler Alert: You actually didn't solve the simultaneous equation. You're meant to reach the final solution analytically and not by estimation or shooting.

Solve
1. 1/X + 1/y + 1/z = 1 find x,y and z


2. 5Loga base 4 + 48log4 base a = a/8.

donedy: Spoiler Alert: Hello bro, I don't want to cause spoiler-alert but you got your solution wrong. You might be lucky on your answer. The properties applied on the shaded equation is invalid: log(ab) = log(a) + log(b, which implies that log(a+b) NotEqual log(a)+log(b). I'm glad you've been corrected by other people. This is a nonlinear equation which can only be solved numerically or by graphical method.

he had already been corrected severally, anyways thanks for the observation and welcome to the thread

@double dy/dx, i dey see ur handwork, u'r a great man, i salute you!

donedy:

Spoiler Alert: You actually didn't solve the simultaneous equation. You're meant to reach the final solution analytically and not by estimation or shooting.

Mr Spoiler alert, it would enlighten us some if you gave us the correct answer.