Solving Polynomial Equations: Help With (x+3)^2/3=3 & 1-1=3

[gfighter39]

i need help with a couple of problems

1.

(x+3)^2/3=3

I don't even know were to start

2.1 - 1 = 3
- ---
x x+1

LCD is (x)(x+1)

1(x)(x+1) - 1(x)(x+1) = 3(x)(x+1)
--------- --------- ---------
x(x+1) (x)x+1 (x)(x+1)

1x+1 - 1x = 3x^2 + 3x

3x^2 + 3x + 1 = 0

3^2 - 4(3)(1)

that is all i have and i don't think it is right because the answer should be

-3 + or - the squ. root of 11
-----------------------------
6

 

[NateTG]

You've got:
$$\frac{(x+3)^2}{3}=3$$
so
$$(x+3)^2=9$$
right?

 

[HallsofIvy]

Assuming that (x+3)^2/3=3 meant $$\frac{(x+3)^2}{3}=3$$ then
NateTG is correct: (x+3)²= 9.

2. is a little hard to understand. I think you meant
$$\frac{1}{x(x+1)}-1= 3$$.

Yes, the LCD is x(x+1) but then you used that incorrectly. If you want to combine fractions on the left, the first fraction remains $$\frac{1}{x(x+1}$$. You don't want the "x(x+1)" in the numerator. You multiply both numerator and denominator of the "-1" by x(x+1) to get $$\frac{1}{x(x+1)}- \frac{x(x+1)}{x(x+1)}= 3$$. The "3" also does not change on the right hand side.

Actually, rather than adding the fractions I would recommend simply multiplying both sides of the equation by the LCD:
$$x(x+1)(\frac{1}{x(x+1)}-1)= 3(x(x+1))$$so
$$1- x(x+1)= 3x(x+1)$$ or
$$2x(x+1)= 2x^2+ 2x= 1$$
Now solve that by completing the square or using the quadratic formula.