Alpha Maximus: Alright I guess this quiz can't end that quickly, with hope, it should even protrude into the night , like OBJ's stomach. Since I can't wait for my battery to DEAD(sorry , na Patience fault), I'll be back with 10 medium questions to serve as an appetizer!!

the first round of the quiz will be starting by 10am 2mao
so start getting your questions and answers ready as a potential quiz master. you'd be briefed on what next to do very soon

Alpha Maximus: When exactly is this quiz meant to commence 'cause its way past 3.00 Prime Meridian already(forgive my EFFIZY ). I don't want this battery to fail me yet, already below half!! * rushes to hospital with phone in hand*

No rushing bro! we are still on the stage of selecting quiz masters!
the whole competition will run from Monday to Friday because different rounds will be involved.
will send all selected quiz masters a briefing soon

INTERESTED...

d citizen:
Solution to 3^x+4^x=5^x

Divide bothsides by 4^x

0.75^x+1=1.25^x

(1-0.25)^x+1=(1+0.25)^x

we are having a case of (1+x)^n= 1+nx as a good approximation since x is small enough to ignore the values of x^2, x^3, etc.

Thus, 1-0.25x+1=1+0.25x
1=0.5x
Thus, x=1/0.5=2.

"I am interested"

@richiez,

I would like to make some recomendations with regards to the quiz competition

1. The levels of the participants should be clearly stated especially, jambite, undergraduate, postgraduate or is the competition opens to all and sundry

2. the questions should be objective based with 4 options, a, b, c and d

3. The questions sets should be adaptive for the potential participants. I mean the question to be set should be graded according to the level of difficulties especially, easy, medium and hard case questions (based on steps, reasoning and math areas)

4. If it is objective based question, negative marking should be strictly applied. That is actual mark = correct - incorrect/4.

5. Rewards should be given to the participant with decorated hall of fame to inspire young and aspiring mathematicians or otherwise.

6. Areas of the mathematics to be cover by the participants need to be made know to encourage speed, accuracy, organisation and preparation.

The strcuture and organisation must be matured, sound and solid to encourage more participants in the nearest future. U never can tell how far the kinds of quiz can go.

Msg from the five star general in the house.

d citizen:
@richiez,

I would like to make some recomendations with regards to the quiz competition

1. The levels of the participants should be clearly stated especially, jambite, undergraduate, postgraduate or is the competition opens to all and sundry

2. the questions should be objective based with 4 options, a, b, c and d

3. The questions sets should be adaptive for the potential participants. I mean the question to be set should be graded according to the level of difficulties especially, easy, medium and hard case questions (based on steps, reasoning and math areas)

4. If it is objective based question, negative marking should be strictly applied. That is actual mark = correct - incorrect/4.

5. Rewards should be given to the participant with decorated hall of fame to inspire young and aspiring mathematicians or otherwise.

6. Areas of the mathematics to be cover by the participants need to be made know to encourage speed, accuracy, organisation and preparation.

The strcuture and organisation must be matured, sound and solid to encourage more participants in the nearest future. U never can tell how far the kinds of quiz can go.

Msg from the five star general in the houe.

'just said my mind'...

I will be following for now,

Five star general approah to 1^x+2^x+3^x=6^x

(6-5)^x+(6-4)^x+(6-3)^x=6^x

(6^x(1-5/6)^x)+(6^x(1-4/6)^x)+(6^x(1-3/6)^x)=6^x

6^x((1-5/6)^x)+(1-4/6)^x+(1-3/6)^x)=6^x

1-5/6x+1-4/6x+1-3/6x=1
3-12/6x=1
2x=3-1
2x=2
X=2/2=1

N.b use the principle of linear approximation((1+x)^n approximately equal to 1+nx) becos x is small enough to ignore the values of x^2, x^3, etc

jackpot:

At the end of 46 days, the snail must have moved
46 * 0.5m = 23m
then, on the 47th day, you will agree with me that the snail will first of all move up 2m before sliding down.
So, 23+2=25m, and the snail is already out!

But if you still argue that even though he has reached the top that he must slide down 1.5m, then on the 48th day, the snail must first of all move up 2m, and this gives
47*0.5+2=25.5m

and this time around, the snail don commot from the top tey tey and no need sliding down.

So dear, I'm sorry but you didn't get the answer right, Sir!!!

hmmm, nice one from you Aunt. A similar stuff asked by a friend of mine: 'Peter climbs a mountain 500m high. Each day he climbs 200m but because he is clumsy and drops things, he goes down 100m the next day. In how many days will he get to the top of the mountain?
Not needing its solution anyway.

Wow, it's Innovative!
Can we still recommend? I recommend; 'calculusf(x).

I'm sorry.i have not been typing for quite a long time,it's just that someone whose life i cherish than mine is sick,so i don't have much time nowadays,i'm very sorry....thanx to master ortarico for remembering calculusf(x) on nairaland maths clinic...and i also greet all generals in the house starting from master richiez,double dx,and my greatest oga here who's the five star maths general(i think everybody knows him with that) master d citizen(who's the only general in the same field with me here),master 2nioshine,master jackpot...and my other generals,starting from master laplace,master alpha maximus,my brother who's also my master(ridex),master labodinho and my great pals who are my masters....benbuks and ortarico....I RESPECT YOU ALL...let's keep the thread rolling...

Ortarico: Wow, it's Innovative!
Can we still recommend? I recommend; 'calculusf(x).

hmmmm

d citizen:
@jackpot,

Pls donot institute a case against me. Am jst being curious about ur true identity and making a honest and passionate appeal. I have gone tru ur incisive post on this forum but at different sections especially joke and romance and i saw a smooth oscillation of ur identity from ferminism to masculinity. I really admire ur math insight and prowess .i must admit that you have a deep seated passion for math. I can see that tru ur indepth scruitiny of the working of math problems by fellow math generals in the house and i am interested in knowing ur true identity. Great female mathematicians are few but they come once in a century. I am putting you on a hotseat and challenge u to review ur identity now. I want my conscience to be clear when referring to ur gender with pronouns "he" or "she". I must salute ur courage, personal drive and sheer determination as you journey into the math world.

honest n unbaised msg from the five star general.

...we are thinking the same master

@calculus

It is a good thing dat u are back. I wish ur loved one a speedy recovery. Keep the math clinic running with ur great and incisive contribution. Take heart and get urself together again

[quote author=Alpha Maximus][/quote]@olarewaju's question on the expansion of (a+b+c)(a+b-c)(a-b+c)(-a+b+c)...re-arrange to give (a+b+c)(a+b-c){c+(a-b)}{c-(a-b)}...check the ones that formes difference of two squares there...then {(a+b)^2-c^2}{c^2-(a-b)^2}...expand to give {a^2+b^2+2ab-c^2}{c^2-(a^2+b^2-2ab}...then,{a^2+b^2+2ab-c^2}{c^2-a^2-b^2+2ab}...when you multiply by -1 and you make it affect the second bracket{a^2+b^2+2ab-c^2}{-c^2+a^2+b^2-2ab}...re-arrange to give {a^2+b^2+2ab-c^2}{a^2+b^2-2ab-c^2}...{a^2+b^2-c^2+2ab}{a^2+b^2-c^2-2ab}...which is another difference of two squares...{[a^2+b^2-c^2]^2-[2ab]^2}...then expand the last to give a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2

Hmmmm

Alpha Maximus: @Calculusf(x) , I've seen , studied and learnt from your work here on this thread and you have proven your worth and skill with numbers time and time again with unique approaches and also that your characteristic humility.Godspeed to your ill acquaintance though and I wish him/her a speedy recovery.

.thank you my brother,i respect you mathematical approaches to question,you are great bro.and try to be greater

d citizen:
@calculus

It is a good thing dat u are back. I wish ur loved one a speedy recovery. Keep the math clinic running with ur great and incisive contribution. Take heart and get urself together again

...thank you very much my master,i'm very grateful.

Factorise x^4 +x^2b^2 +b^4

d citizen:
@richiez,

I would like to make some recomendations with regards to the quiz competition

1. The levels of the participants should be clearly stated especially, jambite, undergraduate, postgraduate or is the competition opens to all and sundry

2. the questions should be objective based with 4 options, a, b, c and d

3. The questions sets should be adaptive for the potential participants. I mean the question to be set should be graded according to the level of difficulties especially, easy, medium and hard case questions (based on steps, reasoning and math areas)

4. If it is objective based question, negative marking should be strictly applied. That is actual mark = correct - incorrect/4.

5. Rewards should be given to the participant with decorated hall of fame to inspire young and aspiring mathematicians or otherwise.

6. Areas of the mathematics to be cover by the participants need to be made know to encourage speed, accuracy, organisation and preparation.

The strcuture and organisation must be matured, sound and solid to encourage more participants in the nearest future. U never can tell how far the kinds of quiz can go.

Msg from the five star general in the house.

NOTED!

The Nairaland math quiz round1 has commenced! Click Here to participate

@ Olarewaju questn on d expansion of (a+b+c)(a+b-c)(a-b+c)(-a+b+c).
Here is anoda approach.
expansn of (a+b+c)(a+b-c)=a^2+b^2-c^2+2ab.
For d expansn of (a-b+c)(-a+b+c)=-a^2-b^2+c^2+2ab.
nw we av (a^2+b^2-c^2+2ab)(-a^2-b^2+c^2+2ab) by expansn we av -a^4-b^4-c^4+2a^2b^2+2a^2c^2+2b^2c^2. Multiplyin thru by -1 we av a^4+b^4^c^4-2a^2b^2-2a^2c^2-2b^2c^2.
All is well.

rhydex 247:
@ laplacian.
Here is the soln.
a^(x^a) implies
(a.x)a-(a.a)x. Where x=(x1,x2,x3) let a=(a1,a2,a3).
a.a=(a1,a2,a3)(a1,a2,a3).
a.a=a^21^a^22+a^23.
a.x=(a1,a2,a3)(x1,x2,x3).
a.x=a1x1+a2x2+a3x3.
a^(x^a)=(a1x1+a2x2+a3x3)a-(a^21+a^22+a^23)x. Recall that a=(a1,a2,a3) nd x=(x1,x2,x3).
a^(x^a)=(a1,a2,a3)(a1x1+a2x2+a3x3)-(x1,x2,x3)(a^21+a^22+a^23). When u expand nd open d bracket u get 0.
Hence a^(x^a)=0. Buh 4rm d questn we av x+a^(x^a)=b. Put a^(x^a)=0 in d eqn we av
x+0=b. x=b. Where x=x1,x2,x3. Hence x=(x1,x2,x3)=b. Note ^ means cap but ^ in a^21+a^22+a^23 means raise to power. All is well.

i'm interested

Wow! It's really great to be here...just notice this thread today...I'm now part of you...keep the ball rolling, will be contributing too,

Kudos!

benbuks: Factorise x^4 +x^2b^2 +b^4

(b^2-bx+x^2)(b^2+bx+x^2).

@benbuck

X^4+x^2b^2+b^4
(x^4+b^4)+(x^2b^2)

((x^2)^2+(b^2)^2)+(xb)^2

((x^2+b^2)^2-2(xb)^2)+(xb)^2

(x^2+b^2)^2-(xb)^2

(x^2+b^2-xb)(x^2+b^2+xb)

Or((x+b)^2-3xb)((x+b)^2-xb)