can someone help me with a little math?

Question

I'm on summer vacation and have forgotten how to do these few problems.

Find all real solutions of
The square root of (x squared - 12x + 96 = 8) and
(x squared - x - 42) divided by (x - 7)

solve: 3(x+4)squared - 162 = 0 and 5x = 3x squared -1

Indicate the Domain of the function
6x divided by x(x squared -49)

if you could explain how it's done as well as give the answer I would be most grateful

this is summer work I can't figure out so I have no textbook to refer to thats why I am here asking for help

Answer 1

do you really want someone else to think for you?
hmmm...well then...
here it goes...
1. look in your math book...there are examples of how to do this problem...
2. same answer as above!!!
#3... same...

#4!!!! remember Einstein solved "un-solve able" problems!!!

Answer 2

1. If the "square root of" applies to the (x squared -12x + 96)
only then we have

(square root (x squared -12x + 96)) = 8

Squaring both sides,

(x squared -12x + 96) = 64
(x squared -12x +32) = 0

Using the quadratic formula,

x= (-b + square root( b squared -4ac))/2a

or

x= (-b - square root ( b squared -4ac))/2a


where a is the coefficient in front of x squared.
b is the coefficient in front of x and
c is the remaining integer.

In this case a=1, b= -12 and c = 32 which
solves to x =4 or x=8.


2. (x squared -x -42) / (x-7)

= (x-7)(x+6) / (x-7)

= (x+6)



3. 3(x+4) squared -162 = 0 implies

3(x+4) squared = 162

(x+4) squared = 54

(x squared + 8x + 16) = 54

(x squared + 8x - 38) = 0


Using the quadratic formula again, x = -4 +3 * square root (6)
or x = -4 - 3* square root(6)

4. 5x =3x squared -1

3x squared - 5x -1 = 0

Using the quadratic formula again,
x= (5+ square root(37)) / 6 or
x= (5 - square root(37)) / 6

5. Domain of 6x/(x(x squared -49))

Here the denominator cannot be zero so values
of x that make the denominator equal zero are
not part of the domain. Those numbers are
7, 0, and -7. So the domain is

x>7
0 -7 x<-7

Hope this helps.

Answer 3

(x^2-12x+96=8)^1/2 doesn't make any sense....
I think you meant (x^2-12x+96)^1/2=8
Square both sides:
x^2-12x+96=64
x^2-12x+32=0
This is a quadratic equation, so the roots are
x=-b+/-(b^2-4ac)^1/2
----------------------
2a
Sorry it looks so ugly here. In words, x equals (minus b plus or minus the square root of (b squared - 4ac)) all over 2a.
a is the coefficient of x^2 in the quadratic equation, b is the coefficient of x and c is the constant. (ax^2+bx+c=0)

There will be two roots, 14 and 2. Please check to make sure you get the same answer.


For the second part, (3x+4)^2-162=0 and 5x=3x^2-1. Write these in the form ax^2+bx+c=0 also. These are what you should get:
9x^2+24x-146=0
3x^2-5x-1=0

Solve using the quadratic equation again.
x=322.6667, -325.3333
x=7, -5.3333


Third, not sure if I'm reading this right:
y=6x
---- ????
x(x^2-49)

If that is right, the answer would be
(-infinity,-7)U(-7,0)U(0,7)U(7,infinity)
If you don't understand my notation, the domain is all real numbers except -7,0, and 7. This is because the denominator will be 0 at those values of x, and as I hope you know, division by 0 is impossible.

Hope this helps--You can e-mail me anonymously if you don't understand anything I said.

Answer 4

Just leave it. Go for a swim or something. It gives me a headache just reading it.

Answer 5

have a rest for god's sake

Answer 6

lol
no we can't do your homework
it would not be ethical

Answer 7

you are bugging