Solving By Factoring
(x-8)(x-5)=x
Please show work!!!
Don't change names, stick to one name
follow the same steps I showed you yesterday for the same type of problem
http://www.jiskha.com/display.cgi?id=1415491412
expand first, then bring everybody to the left and factor.
Remember the right side has to be zero.
What are you talking about?
you just proved my point, you forgot to switch back to one of your other names
To solve the equation (x-8)(x-5) = x by factoring, follow these steps:
Step 1: Expand the equation
(x-8)(x-5) = x
x^2 - 5x - 8x + 40 = x (using the distributive property)
x^2 - 13x + 40 = x
Step 2: Simplify the equation
x^2 - 13x + 40 - x = 0 (subtracting x from both sides)
x^2 - 14x + 40 = 0
Step 3: Factor the quadratic expression
To factor the expression, we need to find two numbers whose product is 40 and whose sum is -14. The numbers are -10 and -4.
x^2 - 14x + 40 = (x - 10)(x - 4) = 0
Step 4: Set each factor equal to zero and solve
Setting (x - 10) = 0, we get x = 10.
Setting (x - 4) = 0, we get x = 4.
Thus, the solutions to the equation (x-8)(x-5) = x are x = 10 and x = 4.