basic algebra and arithmetic "marathon". (1 Viewer)

[Riot09]

I'm pretty sure most of you know what a "marathon" is but for those who don't its basically a game.

in short a person posts a question for the current topic and somebody else solves it and posts their own question related to the topic.

-Please include working out (since most 3u and 2u exams rrquire it and it helps other people learn.Thank you for reading this and i hope you participate.

1.(x+2)^2 + (x-1)(x-7)=17

 

[stevey6404]

(x+2)^2 + (x-1)(x-7) = 17
Now Expand...
x^2 + 4x+ 4 + x^2 - 8x + 7- 17 = 0
Simplify.
2x^2 - 4x - 6 = 0
(2x-6)(x+1) = 0
Therefore x = 3 or x = -1

Okay, my turn.

"In the notes it was proven that x^2 + y^2 >= (large than or equal to) 2xy. Use this result and appropriate substitutions to prove

a) a + (1/a) >= 2, for a > 0
b) (a+b)/2 >= Square root (ab), for a and b both positive.

 

[Riot09]

how come you didnt add a quesion stevey?.Anyway ill just add one from a book.

solve:25x^2=81

 

[stevey6404]

how come you didnt add a quesion stevey?.Anyway ill just add one from a book.

solve:25x^2=81

Oh i forgot :L
but I edited it but I'll post it again after this answer:

25x^2 = 81
Square root both side..imagine it as Square root 25 x square root x^2 = square root 81
5x = +- (positive negative) 9
therefore x = 9/5 or -9/5

"In the notes it was proven that x^2 + y^2 >= (large than or equal to) 2xy. Use this result and appropriate substitutions to prove

a) a + (1/a) >= 2, for a > 0
b) (a+b)/2 >= Square root (ab), for a and b both positive."


If you can't solve it, just post a new question. I'll just repeat this question after every post.

 

[Riot09]

okay,im pretty new to this-are you sure this is in this topic?,anyway i think iv'e got an answer.

a) a + (1/a) >= 2, for a > 0

therefore is a=2 it follows that 2+2/1 >2 (a>0)
2.5 > 2
therefore the equation is correct.

b)(a+b)/2 >= (ab),Square root for a and b both positive.

therefore 2+2/2 >or equal to Square root of 2x2

4/2 >oe equal to Square root of 4

2>or equal to 2

therefor the equation is correct

(i havent done this type of question before so can you please tell me where you got it from,which txtbook or was it in a exam,classnotes etc.

 

[muzeikchun852]

since u havent post any question.
i will post one:

|2x-1| = |x+4|

 

[stevey6404]

since u havent post any question.
i will post one:

|2x-1| = |x+4|

|2x-1| = |x+4|

(2x-1)^2 - (x+4)^2 = 0
difference of 2 squares
[(2x-1)-(x+4)] [(2x-1)+(x+4)] = 0
(x-5)(3x+3)=0
therefore x = 5 or x = -1

there's also another way to do it but it's longer.

 

[ohexploitable]

Since no one posted a new question...

 

[x jiim]

In the notes it was proven that x^2 + y^2 >= (large than or equal to) 2xy. Use this result and appropriate substitutions to prove

a) a + (1/a) >= 2, for a > 0
b) (a+b)/2 >= Square root (ab), for a and b both positive."

okay,im pretty new to this-are you sure this is in this topic?,anyway i think iv'e got an answer.

a) a + (1/a) >= 2, for a > 0

therefore is a=2 it follows that 2+2/1 >2 (a>0)
2.5 > 2
therefore the equation is correct.

I think it is in the topic- sub x=sqrt(a) and y=1/(sqrt(a))
(sqrt(a))^2 + (1/(sqrt(a)))^2 >= 2*(sqrt(a))*(1/(sqrt(a)))
a + 1/a >= 2

b)(a+b)/2 >= (ab),Square root for a and b both positive.

therefore 2+2/2 >or equal to Square root of 2x2

4/2 >oe equal to Square root of 4

2>or equal to 2

therefor the equation is correct

(i havent done this type of question before so can you please tell me where you got it from,which txtbook or was it in a exam,classnotes etc.

Sub x=sqrt(a) and y=sqrt(b)
(sqrt(a))^2 + (sqrt(b))^2 >= 2*sqrt(a)*sqrt(b)
a + b >= 2sqrt(ab)
(a+b)/2 >= sqrt(ab)

that's how you do those questions, used to get them all the time in class and exams etc. it's useful to know how to do them in case you get asked.

 

[jet]

Hi guys, there is already a 2 unit marathon thread in the 2 unit mathmatics forums. It can be found here: http://community.boredofstudies.org/12/mathematics/234357/2-unit-revising-marathon-hsc-10-a.html

 

[stevey6404]

I think it is in the topic- sub x=sqrt(a) and y=1/(sqrt(a))
(sqrt(a))^2 + (1/(sqrt(a)))^2 >= 2*(sqrt(a))*(1/(sqrt(a)))
a + 1/a >= 2

Sub x=sqrt(a) and y=sqrt(b)
(sqrt(a))^2 + (sqrt(b))^2 >= 2*sqrt(a)*sqrt(b)
a + b >= 2sqrt(ab)
(a+b)/2 >= sqrt(ab)

that's how you do those questions, used to get them all the time in class and exams etc. it's useful to know how to do them in case you get asked.

omg genius!
How did you know what to let x and y be? Do you just look at it and just get it magically?

Thanks very much for the help =]

 

[ohexploitable]

Hi guys, there is already a 2 unit marathon thread in the 2 unit mathmatics forums. It can be found here: http://community.boredofstudies.org/12/mathematics/234357/2-unit-revising-marathon-hsc-10-a.html

Yeah but we haven't learnt that yet!

 

[stevey6404]

Since no one posted a new question...

(x+2)^2 + (y-1)^2
= x^2 + 4x + 4 + y^2 - 2y + 1
Rearrange.
=x^2 + y^2 + 4x - 2y + 1 + 4
since x^2 + y^2 + 4x - 2y + 1 = 0
= 0 + 4
= 4


My question
Prove that x^2 + xy + y^2 > 0 for any non-zero values of x and y

 

[ohexploitable]

Prove that x^2 + xy + y^2 > 0 for any non-zero values of x and y

Don't know. Answer plz

 

[Absolutezero]

Prove that x^2 + xy + y^2 > 0 for any non-zero values of x and y

Is it meant to be 2xy?

 

[stevey6404]

Is it meant to be 2xy?

Nope, it's xy. It's page 78 question 23, if you have the cambridge 3U book.

 

[t00l]

Nope, it's xy. It's page 78 question 23, if you have the cambridge 3U book.

i got this question for homework . working on it now. damn hard

 

[carmot]

hmm, not too sure about my setting out but:

x^2 + xy + y^2 = x^2 + 2xy + y^2 - xy

= (x+y)^2 - xy

since (x+y)^2 > xy

x^2 + xy + y^2 > 0

my question:

Find the values of k for which the quadratic equation
x^2 - (k+3)x + (k+6) = 0
has real roots

 

[ohexploitable]

my question:

Find the values of k for which the quadratic equation
x^2 - (k+3)x + (k+6) = 0
has real roots

My Question:

 

[Drongoski]

since u havent post any question.
i will post one:

|2x-1| = |x+4|

Another way (is this longer?)

.: 2x-1 = (x+4) or 2x-1 = -(x+4)

Solving, easily:

x=5 or x=-1