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Well, CU......your hint was a GREAT one   !!!

I wouldn't have seen how to proceed without it  !!!

Yes, this is correct!

Using CU's hint :

(p + q+ r)  + (qx + ry)  + (pz + rx)  + ( py + qz)  =  x + y + z

-3   +  x (q + r)  + y (p + r) + z(p + q)  =  x + y + z

-3   + x(-3 - p)  + y (-3 - q)  + z (-3 - r)  = x + y + z

-3  - 3x  - px  - 3y - qy  - 3z - rz  =  x + y + z

-3 - 3 ( x + y + z)  -  ( px + qy + rz)  = x + y + z

-3  - 3(x + y + z)  -  (1)  = x + y + z

-3 - 1  =  4 (x + y + z)

-4  =  4 (x + y + z)

-1  = x + y + z

your close!

 rearrange and subtitute in -3

(qx + ry + pz) + (rx + py + qz) - 3 = x + y + z

Oops I realized the next step is wrong! The problem may be harder than I thought!

-3 ( x + y + z) + (-3) (x + y + z) - 3 = x + y + z           (undistribute p + q + r because they = -3)
 

-3x - 3y - 3z -3x - 3y - 3z - 3 = x + y + z

-6x - 6y - 6z - 3 = x + y + z

7x + 7y + 7z = 3

answer is 3/7

I think so

ik nobody is gonna see this but like holy moly my school REALLY loves tests...i have 3 tomorrow and I cannot concentrate at all oh my lord

Thanks for that hint, CU......that's pretty crafty   !!!

You can ask for the explanation and solution if you are having trouble.

Here is a hint:

See the equations that have isolated x, y, z?

Add them up.

Now, notice the other equations, how can you susbtitute in and evualate?

Its pretty simple, you just have to see it!

thank you Rom <3 

I think I did well, but i think i also made careless errors too :(( 

Holy cow rom that is smart

I always thought that your profile pic was a cow until I looked closely and it was a horse, er, or maybe a pony.

If   the vertex  is  (3,2)    and  it contains the point   (1,0).....then x  = 1 is a  root

So...by symmetry......(5, 0) is also a root

So....we can find  "a"  thusly

0  =  a(1 - 3)^2  + 2

-2  =  a(-2)^2

-2  = 4a

a  =  -2/4  =  -1/2

And by Vieta......

The sum of the roots   -b/a

So

1 + 5  = - b / ( -1/2)

6 = 2b

b = 3

And the product of the roots  = c/a

So

5*1  =  c /(-1/2)

5 = -2c

c = -5/2

So

(a,b , c)   =  ( -1/2, 3, -5/2)

Here's the graph :  https://www.desmos.com/calculator/yd99xbrpsg

\(\left(x+\dfrac 1 x\right)^3 = \\ x^3 + 3x+ 3x^{-1}+x^{-3} = \\ 52 + 3\left(x + \dfrac 1 x\right)\\ u=\left(x+\dfrac 1 x\right)\\ u^3 - 3u - 52=0\\ (u-4) \left(u^2+4 u+13\right) = 0\\ \text{The left hand factor is zero at $x=4$, the right hand factor has complex roots}\\ \text{The solution you are after is $x=4$}\)

Reflect the x intercept over the line of symmetry in the quadratic.

The vertex is (3,7), so the line of symmetry must be x = 3

We reflect that, so 3-(-2) = 5

3 + 5 = 8

So the x coordinate of the other x intercept is (8, 0)

Parabolas are symmetric.....the other zero will be the sam distance from the vertex as the one given...

from -2 to 3 is five units.....the other zero should be at   8,0

We can solve this  for  "a"

-3  = a (-4 -2)^2  + 1

-3 = a ( -6)^2 + 1

-4  = 36a

a  = -4/36  = - 1/9

So....we have this form

y  = (-1/9( ( x - 2)^2  + 1       to find the zeroes....let y  = 0

0  = (-1/9)(x - 2)^2 + 1

-1  = (-1/9)(x - 2)^2       multiply through by -9

9 =  ( x - 2)^2        take both roots

±√9  = x - 2

±3  =  x - 2    add 2 to both sides

±3 + 2  =  x

So....the roots are

x =  3 + 2  = 5  = m         and   x  =  -3 + 2  =  -1  = n

So

m - n  =   5 - (-1)   =  6

LOL!!!!!......

x + 1/x  = 7        square both sides

(x + 1x) (x + 1/x)  = 49

x^2  +  x (1/x)  + x (1/x)  +  1/x^2  = 49     

x^2  +  1  + 1  + 1/x^2  =  49     subtract  1 from both sides and rearrange the left side

x^2 + 1/x^2  + 1  =   48    Voila  !!!!

I graphed it straight from the original values but if you want to divide by something to make it graphable by hand why do you need anyone to do it for you.  There is a calculatoro on the site.

At the y-intercepts , x =0   so set the equation = 0 and solve for y

0 = -y^2+4y-4      use Quadratic Formula to find 'y' = 2    (only one root at y=2)  so only ONE y=intercept

Alternatively , the discriminant    b²-4ac = 0  

                                                   4²-4(-1)(-4)= 0      so only ONE root (One y intercept)     

Rearrange the given equation as

x^2 - 6x + y^2 + 8y  = 24     complete the square on x and y

x^2 - 6x + 9  + y^2 + 8y + 16  =  24  + 9 + 16

(x -3)^2  + (y + 4)^2  =  49

The center of the circle  is  ( 3, -4)

So.....the distance between the center  and (-3, -12)   is given by :

sqrt [ (-3 - 3)^2  + (-12 - - 4)^2 ] =

sqrt  [ (-6)^2  + (-8)^2 ] =

sqrt [ 36 + 64 ] =

sqrt (100) =

10 units

We can solve this by inspection. Logic

If  abc  =  21     and    a + 1/b  =  22/7

Then this implies that

a + 1/b  =  3  + 1/7

Which implies that  a  = 3    and  b  = 7

So

abc  =21

(3)(7)(1)  = 21

So  c  = 1

And

c + 1/a  =    1   + 1/3