You can put this solution on YOUR website! Given
(1) 32 - 32x = (6x+4)(x-1)
Let's do a little factoring on (1). Factor out 32 on the left side and (-2) on the right side to get
(2) 32(1-x) = -2(3x+2)(1-x)
Now divide both sides of (2) by -2(1-x) to get
(3) -16 = 3x + 2 or
(4) 3x = -18 or
(5) x = -6
Let's use (1) to check the value of x.
Is (32-32(-6) = (6*(-6)+4)*(-6-1))?
Is (32 + 192 = (-36+4)*(-7))?
Is (224 = (-32)*(-7))?
Is (224 = 224)? Yes
Answer: x = -6
My solution technique comes from experience, i.e. look for simplifications via common factors. When I looked at (1) I saw that the left side is -32(x-1). So the factor (x-1) will "cancel out" and reduce the equation to a first order.
The "brute force" way to solve (1) is to multip;y all factors and reduce them to a single expression equal to zero. Let's do it.
(6) 32 -32x = 6x^2 -6x +4x -4 or
(7) 6x^2 +32x -6x +4x -32 -4 = 0 or
(8) 6x^2 +30x -36 = 0 or
(9) x^2 +5x -6 = 0 or by factoring we get
(10) (x+6)*(x-1) = 0 or
we get TWO roots
(11) x = {-6,1}.
What happened? Where did I go wrong in the first method? Do you see where I erred?
My simplification resulted in the -6 root but where did the x = 1 go?
Remember when I cancelled out (x-1) by division? Well I was dividing by zero! That's a no no. So be careful taking short cuts. The clue to my error is when a quadratic equation became a first order equation.
Let's use (1) to check if x = 1.
Is (32-32(1) = (6*(1)+4)*(1-1))?
Is (32 -32 = (10)*(0))?
Is (0 = 0)? Yes
What I should have written in (2) is
(12) (1-x)*(32 +2(3x+2)) = 0 or
(13) (1-x)*(32 + 6x +4) = 0 or
(14) (1-x)*(6x + 36) = 0 or
(15) (x-1)*(x+6) = 0 which is the same as (10)
Moral of the story: Stick to the brute force technique!
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