Solving the Equation (I did something wrong)

[sw87]

I thought I sloved this problem and realized it can't be right. The problem is 5x(x+1)=(4x-3)(x+1) I divided both sides by (x+1). Then divided both sides by 5 and thought the answer was x=(4x-3)/5 this can't be right. What did I do wrong? Please Help

 

[mmm4444bot]

Dividing both sides by x + 1 is not recommended, unless you know ahead of time that x + 1 is not zero.

Before you can isolate x on one side (i.e., solve), you first need to get rid of the parentheses.

Start by multiplying everything out, on each side. You can then collect all terms to one side, for a quadratic equation in general form.

 

[sw87]

The final answer I got came to x=-1/3

 

[mmm4444bot]

The final answer I got came to x=-1/3 That value does not lead to a true statement, when substituted for x in the original equation.

So, it's not a solution.

I won't be able to tell you what you're doing wrong, unless you show me what you've done.

 

[sw87]

5x(x+1)=(4x+3)(x+1)
-(5x(x+1)) -(5x(x+1)) (4x+3)(x+1) foil
0=-5x^2-5x+4x^2+4x-3x-3
0=-x^2-4x-3
0=-x^2-2x^2-3
0=-3x^2-3
0=-9x-3
+9x +9x
9x=-3
divide both sides by 9
x=-1/3

 

[mmm4444bot]

5x(x+1)=(4x+3)(x+1)
-(5x(x+1)) -(5x(x+1)) (4x+3)(x+1) foil
0=-5x^2-5x+4x^2+4x-3x-3
0=-x^2-4x-3 Looks good, up to this point.
0=-x^2-2x^2-3 How did you change -4x into -2x^2 ?

-x^2 - 4x - 3 = 0

Some people like a positive leading coefficient, although it's not required to solve. Multiply both sides by -1.

x^2 + 4x + 3 = 0

Can you solve either of these quadratic equations?

 

[sw87]

I thought 4x is a perfect square and then I tried adding the 2 squares.

so from 0=x^2+4x+3
0=x^2+1x+3x+3
0=x(x+1)+3(x+1)
0=(x+3)(x+1)
set each equal to 0
x+3=0 x+1=0
-3 -3 -1 -1
x=-3 x=-1
x={-3,-1}

 

[mmm4444bot]

I thought 4x is a perfect square Nope. The factor of 4 is a perfect square, but x to the first power is not a square.

The expression 4x^2 is a perfect square because it's (2x)^2.

x={-3,-1} This is correct.

 

[sw87]

Thank You Very Very Much!