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x^2  +  18x  = 27       complete the square on x

x^2 + 18x  + 81  =  27 + 81

(x + 9)^2  =  108       take both roots

x +9  =   √108

x + 9   =  √[ 4 * 9 * 3]

x + 9  =  √4 *√9 * √3

x + 9   =  2 * 3 * √3

x + 9  =  6√3        subtract 9 from both sides

x  = 6√3  - 9

This doesn't have the specified form, Logic....but taking  "a"  as 3 and "b" as - 9  

a + b   =    3 + (-9)   =   -6

honestly i would go to google if i could

other than that i'm like you i'm stumped

lol I kept searching it up but I kept looking at the same one which was wrong thanks CPhill.

Try this, BC  :  https://web2.0calc.com/questions/help-plz_94

CPhill plz try

x^2  + 1300x  + 1300    complete the square on x

Take 1/2 of 1300 = 650.....square it  = 422500

Add it and subtract it

x^2  + 1300x  + 422500 + 1300 - 422500  

Factor the first three terms and simplify

(x + 650)^2  - 421200

c = - 421200     b  = 650

c  /  b   =    - 421200  / 650   =   -648

aops does not always work for me so i do mine on paper or on a text formate like this forum so?

We can see if we can find a solution for this system

a   +  b   - c    =  0       (1)   

2a  + 4b + 6c  = 0       (2)

a    + 5b  + 15c = 0     (3)

Multiply  (1) by 6   and add to (2)  and we get that

8a + 10b  = 0    (4)

Multiply (1) by 15 and add to (3)

16a + 20b = 0    (5)

(5)  is a multiple of (4).....so these are dependent

8a  = -10b

a = -10/8  b

a = -5/4 b

b = -4/5a

And 

a + b - c  = 0

a - (4/5)a  - c  = 0

(1/5)a  =  c

So...if we let a  =  5    then  b = (-4/5)(5) = -4   and c = (1/5)(5)  = 1

Check

5 ( 1, 2, 1)   - 4 ( 1, 4 , 5 )  + 1 ( - 1, 6 , 15 )   =  ( 0 , 0 , 0 ) 

                          x + 4    

f(x)    =      ___________

                   x^2 + ax + b

If we have vertical asymptotes  at  x  = 1  and  x  = -2  then the denominator  is  ( x - 1) ( x + 2)

So.....the denominator  becomes  x^2 - 1x + 2x - 2  =   x^2 + 1x  - 2

So     a  =  1   and b =  - 2

And their sum is   - 1

x(x + 5)  = -n

x^2 + 5x  = -n

x^2 + 5x  + n  = 0

This will have no real solutions  when

5^2  - 4(1)n < 0

25 - 4n < 0

25  < 4n

25/4 < n     this means that this will have no real solutions when

n > 6.25

So...it will have no real solutions when  n  =  7, 8, 9  or 10

So....the probability of no real solutions =  4 /10   =    2 / 5

Here :   https://web2.0calc.com/questions/help-help-quick-quick

ok..thank you and sorry for the trouble

I don't know of any definition, Manuel......the locus of the points will lie on the diagonal

any definition?

THX, Guest....correction made and my answer agrees with yours!!!

Look at the image here, Manuel :

Notice that every point on the diagonal  BD will be equidistant from AB  and BC

Sorry, the answer was incorrect, here's the solution:

\(\text{Multiplying both sides of the equation by $3jk$ to clear the denominator gives $3k + 3j = jk$. Re-arranging and applying Simon's Favorite Factoring Trick, it follows tha t $$jk - 3j - 3k + 9 = (j-3)(k-3) = 9.$$ Thus, $j-3$ and $k-3$ are pairs of positive factors of $9$, so $(j-3,k-3) = (1,9),(3,3),(9,1)$. These give $k = 4,6,12$, and their sum is $4 + 6 + 12 = \boxed{22}$.}\)

I'd use this instead  :

https://www.desmos.com/

It's an excellent grapher   !!!

Thanks!!! 

Just use the ones on your keyboad ....the ones you used to ask your question. ...   ~EP

Here's a trick I learned from one of our excellent  answerers  (heureka)

Let  3  = z

Let j = z + a

Let k =  z + b   .....so we have

  1                   1                     1

_____   +     ______   =   ______      get a common denominator on the left

z + a             z + b                z

[ z + b ] + [ z + a ]                   1

________________   =       ____

 z^2 + az+ bz + ab                  z

2z + b + a                                  1

_________________  =      _______        cross-multiply

z^2 + az + bz + ab                    z

2z^2 + az + bz  =  z^2  + az + bz +  ab

2z^2   = z^2  +  ab

z^2   = ab

Since z  = 3

Then

ab  =  3^2

ab  = 9

So....we are looking for all the pairs of positive integers  that multiply to 9

So we have

a      b

1      9

3      3

9      1

So  since  j  =  z + a       and   k =  z + b

Then we have

j        k

4      12

6      6

12    4

So.....the sum of all possible values for k   =  22

EDIT TO CORRECT AN ERROR....

Oops, I meant \(k\), not \(K\). And positive, not positize. Sorry.

Correct me if im wrong CP but if im right the post an emoji