first line: 35-x^2+2x=0
next important line: x^2-2x-35 = 0 , I don't know why you did all that other stuff in between.
(see my response to a similar problem in the reply to "Lori" two postings down from here)
so x^2-2x-35 = 0
(x-7)(x+5) = 0
then x=7 or x = -5
you had as factors (x-7)(x+5) (x-7)(x+5).
Why 4 of them?? It was only a quadratic, so it could have at most two factors.
that was from the example we were given, I just followed it. Our instructor was ill and the sub handed out examples, no one understood and the sub could not explain! Sorry I have the correct answers, but I need to go back and show my work the right way? I will look in my book. Thanks
Here is all you would have needed, down to the essential steps
35-x^2+2x=0 (multiply by -1 to get it in standard form
x^2 - 2x - 35 = 0 (this factors to)
(x-7)(x+5) = 0
therefore x-7 = 0 or x+5 = 0
therefore x = 7 or x = -5
If you are asked to verfy the solution, this is how you would do that:
if x=7, then
L.S. = 7^2 - 2(7) - 35
= 49 - 14 - 35 = 0
if x= -5
L.S. = (-5)^2 - 2(-5) - 35
= 25 + 10 - 35
Therefore x = 7 or x = -5
Algebra - Solving quardratic equations using factoring. 2X^2+2X-4=0 The book ...
Algebra-Graphing Linear Equation - Y<2X+4 The way I unerstand from the book ...
Math - 2x(x+3) = 0 Is there a way to solve that using the below methods? ...
algebra 2 - solve this equation by factoring a^2-8a=-15 the a^2 is a' squared
Algebra - Use factoring to solve the equation: x^2+13x+36+0
Algebra - 6x^2+9cx= 6c^2 I tried moving the 6c^2 to the other side of the ...
algebra - Write the equation in quadratic form and solve it by factoring. x2(8x...
Algebra - Steps to factoring this x^3 - 8 ? I'm only used to factoring out the x...
Algebra - Solve the equation by factoring: 6y^2+7y-3=-12y^2-54y+4 The answers ...
Algebra - I am working on Factoring the Trinomial and need help to solve this ...