SOLUTION: I'm having issues with these types of problems sinking in. -5x+3=x^2+2x, -5x+3-x^2+2x=0 then we're left with -2x+3-x^2 we rearrange that to look like -x^2-2x+3 then...x^2+2x-3 (x+

Algebra ->  Rational-functions -> SOLUTION: I'm having issues with these types of problems sinking in. -5x+3=x^2+2x, -5x+3-x^2+2x=0 then we're left with -2x+3-x^2 we rearrange that to look like -x^2-2x+3 then...x^2+2x-3 (x+      Log On


   

Question 1036813: I'm having issues with these types of problems sinking in. -5x+3=x^2+2x, -5x+3-x^2+2x=0 then we're left with -2x+3-x^2 we rearrange that to look like -x^2-2x+3 then...x^2+2x-3 (x+3)(x-1) I feel like I'm doing it wrong. Here's the problem.....
Find the values of x for which f(x) = g(x)
a. f(x) = -5x + 3, g(x) = x^2 + 2x

Found 3 solutions by josgarithmetic, addingup, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
-5x%2B3=x%5E2%2B2x

First use basic algebraic properties.
-5x%2B3-x%5E2-2x=0
-5x-2x-x%5E2%2B3=0
-7x-x%5E2%2B3=0
-x%5E2-7x%2B3=0
%28-1%29%28-x%5E2-7x%2B3%29=%28-1%29%2A0
highlight_green%28x%5E2%2B7x-3=0%29------not factorable in integer coefficients.

You can solve for the variable using either complete-the-square or going directly to general solution formula. (Quadratic Equation)

Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = -5x+3, g(x) = x^2+2x Move all the terms to the left side:
x^2+2x-5x+3 = 0 Combine like terms:
x^2-3x+3 = 0
Now we have a quadratic equation in the form ax^2+bx+c, so i'll solve it by using the quadratic formula:
x = -(-3÷2)+-√(-3÷2)2-3
Run the calculations and you get:
x = 1.5±0.866025403784i This calculation has no real solution. Usually you don't have to go beyond this point.

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
I'm having issues with these types of problems sinking in. 
-5x+3 = x^2+2x, 

-5x+3-x^2+2x = 0                       ! Wrong.  Correct is:  -5x+3-x^2-2x = 0

then we're left with -2x+3-x^2         ! Naturally, all that follows is wrong . . . 

we rearrange that to look like 

-x^2-2x+3 then...x^2+2x-3 (x+3)(x-1) 

I feel like I'm doing it wrong. Here's the problem.....

Find the values of x for which f(x) = g(x)
a. f(x) = -5x + 3, g(x) = x^2 + 2x
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Your problem is in that you don't know the rule of moving terms from one side of an equation to the other side.
The rule is: at each such move you MUST change the sign of the term to the OPPOSITE one.