Simple Math Practice Problems

[Dnas]

last samurai is correct.

Explanation

x^4+x^2+1=

x^4+2x^2+1-x^2= (add and subtract x^2)

(x^2+1)^2-x^2= ((a+^2=a^2+2ab+b^2)

(x^2+x+1)(x^2-x+1) (a^2-b^2=(a+(a-)

In case anyone wants to know how to do my other one,

A polynomial of degree 4, in the form x^4+ax^3+bx^2+cx+d=0 (a,b, and c are all real numbers), has roots -4+3i and 9+4i. What is the value of a?

For a polynomial to have real coeficcients and still have complex roots, each complex root must be paired with its conjugate. So, if two roots are -4+3i and 9+4i, the other two must be -4-3i and 9-4i.

In any polynomial in the form,

ax^n+bx^(n-1)+cx^(n-2)+dx^(n-3)+ex^(n-4)...

-b/a is always the sum of the roots of the equation.

So, -a/1=-a= (-4+3i)+(9+4i)+(-4-3i)+(9-4i)=10

So, a = -10

[CodeOptimist]

Nice explanation, Dnas

[CodeOptimist]

If last_samurai doesn't ask a new question in a little while, I'll post up a new one

[akya]

Akya - Oh well, if nobody else is going to guess:

f(x,y) = x^2 + 4xy + sin y + y^2

(df/dx)y = 2x + 4y

(df/dy)x = 4x + cos y + 2y

yep got it right...^^ sorry I didn't answer immediatly

how about this :

Find a polar equation for the circle x^2+(y-3)^2=9

[King Tutankhamun]

Hey that is so cool, we just learned that equation today!

The radius is 3, the center is (0, 3) I think.

Is that what you are asking for?

[MarkT]

Polar-cartesian substitutions

x = r * cos( t )

y = r * sin( t )

x ^ 2 + ( y - 3 ) ^ 2 = 9

r^2 * cos(t)^2 + r^2 * sin(t)^2 - 6 * r * sin(t) + 9 = 9

r^2 * ( cos(t)^2 + sin(t)^2 ) - 6r*sin(t) = 0

r^2 - 6r*sin(t) = 0

r = 6 * sin(t) for 0 <= t < pi/2

[King Tutankhamun]

Humm, I guess that I thought it was too simple.

[CodeOptimist]

Humm, I guess that I thought it was too simple.

Hehe, I don't blame you... Did Mark get the correct answer?

[akya]

yep his turn to ask a Q I guess

[CodeOptimist]

Tick tick... Mark, where art thou?

[King Tutankhamun]

Anybody can feel free to ask the next one if ever someone does not relpy within a while.

[Curufinwe]

(

okay, factorize :

A= (-25 + 9x²) - (25 + 9x² - 30x)

[Dnas]

30x-50 = 10(3x-5)

[chichigrande]

A= (-25 + 9x²) - (25 + 9x² - 30x)

-30x - -50 -> -30x + 50 -> -30x = -50 -> x = 5/3

Did I do it right?

[last samurai]

A=10(3X-5)

[Argalius]

I got this:

A= (-25 + 9x²) - (25 + 9x² - 30x)

A= 9x² - 50 + 9x² - 30x

A= - 50 - 30x

A= 10(- 5 - 3x)

A= 10(- 3x - 5) or just - 30x - 50

[Curufinwe]

(

These are all correct ...

A = 10(3x-5)

Dnas can ask a new question.

[Argalius]

What did I do wrong?

[Dnas]

I've attached it as an image.

Simplify.

Note: Please show some sort of work to prove that you didn't use a calculator.

If this is too hard, I'll post a different one.

[last samurai]

that is way nuts, too hard

[akya]

dunno if I got it

[Dnas]

Nope.

I'll give a few hints, then if needed, I'll change it.

First hint:

Try cubing it.

[Mystic-Al-Bob]

@ akja: nice try but you can't just connect the roots

I also think cubing is a good idea

[Argalius]

I have no idea what cubing is but this is what I think:

2³√2, can't think of anything else.

[MarkT]

Deep breath... and:

A = (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3)

A^3 = (2+sqrt(5)) + 3.((2+sqrt(5))(9-4.sqrt(5)))^(1/3) + 3.((2-sqrt(5))(9+4.sqrt(5)))^(1/3) + (2-sqrt(5))

= 4 - 3.[ (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3) ]

= 4 - 3.A

A^3 + 3.A - 4 = 0

(A-1)(A^2+A+4) = 0

A = 1. I'm not going to bother with the imaginary roots.

Good enough?