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# i need help (solved!)

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ty to the guy who answered!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018

#1
+1

Here's a few

y = x            (linear)

Sub  the   second equation into the first for y

x  = x^2       subtract x from  both sides

0  = x^2  - x

x^2  - x   =  0   factor

x ( x - 1)  =0

Set each factor to 0  and solve for  x

x  = 0        x  - 1  =  0

x = 0          x  = 1

So....when x  = 0, y  = 0

And when x  = 1, y  = 1

So....the solutions are  (0,0)  and (1,1)

Here's a graph ;

https://www.desmos.com/calculator/ldpimov7kd   May 2, 2018
#3
0

y  =   (x + 3)

______

(x - 2)

The horizontal asymptote  occurs at  y  = 1

The vertcal asymptote occurs at  x  = 2

The x intercept is  at  x  = -3

The y intercept is  at   (-3/2)

Here's a  graph :

https://www.desmos.com/calculator/4l3egciy58   May 2, 2018
#5
+1

Let  the polynomial  be    x^2  - 4x  + 4

Let the  linear binomial be    ( x - 2)

Part 1

x  -  2

x - 2   [   x^2  -  4x  + 4  ]

x^2   -2x

_____________

-2x   + 4

-2x   + 4

_______

0

Part 2

f(a)   = f(2)  =   2^2  - 4(2)  + 4   =    4 - 8  + 4   =  0

Part 3

Since  we obtained a "0"  with polynomial long division......then x  = 2   is  a  root. The Remainder Theorem says that if f(a)  = 0, then  "a" is a root.  This is borne out by the fact that f(2) = 0  where  "a"   = 2.   May 2, 2018
#6
+1

Let  N  = 37

We can write this as  (x + y)  =  (49-  12)

So....note that

(49 -  12) ^2  =    49^2  - 2(49*12) + 12^2  =   2401  - 1176  + 144    = 1369  =  372   May 2, 2018
#7
+1

helpful, thank you, you helped me learn a bit too :D

Guest May 2, 2018
#8
+1

x + a      =    b

____           ___

ax               x

We can "engineer"  an extraneous solution, as follows :

Cross-multiply ;

x (x + a)  =  bax    simplify

x^2  + ax  = bax

x^2  + ax - bax  = 0              let  a  =  1, let  b  = 5....so we have

x^2  + 1x  - (5)(1)(x)  = 0

x^2 + 1x  - 5x  = 0

x^2  - 4x  = 0      factor

x(x  - 4)  = 0

Set both factors to 0   and solve for  x

x  = 0           x  - 4  =0

x  = 4

So.....note that the original problem becomes ;

x + 1            5

____   =      __

1x              x

Note that if  x  = 4

4 + 1             5

_____   =    ___       which is true

1*4             4

However...if x  =  0   ....both denominators are undefined because we cannot divide  by 0....thus....x  = 0  is an extraneous solution....!!!!!   May 2, 2018