# Solve for x : 1/(x^2+6x+9) + 1/(x+3) = 4/9

### 4 Answers | Add Yours

1/(x^2 + 6x+9) + 1/(x+3) = 4/9

Let us rewrite:

x^2 + 6x + 9 = (x+3)^2

==> 1/(x+3)^2 + 1/(x+3) = 4/9

=> (1+ (x+3)/(x+3)^2 = 4/9

==> (x+4)/(x+3)^2 = 4/9

Not cross multiply:

==> 4(x+3)^2 = 9(x+4)

==> 4x^2 + 24x + 36 = 9x + 36

==> 4x^2 + 15x = 0

Now factor x:

==> x(4x + 15) = 0

**==> x1= -0**

**==> x2= -15/4**

To solve:

1/(x^2+6x+9) +1/(x+3) = 4/9

Rewrite as:

1/(x+3)^2 +1/(x+3) = 4/9

Multiply by 9(x+3):

9{1 +(x+3)} = 4(x^2+6x+9)

9+9x+27 = 4x^2+24x+36

9x+36 = 4x^2+24x +36

4x^2+24x-9x = 0

x(4x+15) = 0

**x = 0**

**4x+15 = 0**

4x = -15

**x = -15/4**

We notice that the denominator of the first ratio is the result of expanding the square (x+3)^2.

We've applied the formula:

(a+b)^2 = a^2 + 2ab + b^2

So, x^2+6x+9 = x^2+2*3*x+(3)^2

We'll re-write the equation:

1/(x+3)^2 + 1/(x+3) = 4/9

The least common denominator is 9(x+3)^2.

We'll multiply the first ratio by 9, the second ratio by 9(x+3) and the third ratio by (x+3)^2.

We'll re-write the equation, without denominators.

9 + 9(x+3) = 4(x+3)^2

We'll remove the brackets from the left side and we'll expand the square from the right side:

9 + 9x + 27 = 4x^2 + 24x + 36

We'll subtract 4x^2 + 24x + 36 both sides:

9 + 9x + 27 - 4x^2 - 24x - 36 = 0

We'll combine like terms:

- 4x^2 - 15x = 0

We'll factorize by -x:

-x(4x + 15) = 0

We'll set each factor as zero:

-x = 0

x = 0

4x + 15 = 0

We'll subtract 15:

4x = -15

We'll divide by 4:

x = -15/4.

**The solutions of the equation are: {-15/4 ; 0}.**

Here we have to solve for x. From the equation provided:

1/(x^2+6x+9) + 1/(x+3) = 4/9

we can write this as

=> [1 / (x+3)]^2 + 1/ (x+3) = 4/9

Put Y = 1/ (x+3 )

=> Y^2 + Y = 4/9

=> 9Y^2 + 9Y - 4 = 0

=> 9Y^2 + 12Y - 3Y -4 =0

=> 3Y ( 3Y + 4) - 1 ( 3Y + 4 )=0

=> (3Y -1)(3Y + 4) =0

Y = 1/3 and Y = -4/3

As Y = 1/ (x+3)

1/ (x+3) = 1/3

=> x+3 = 3 => x =0

1/ (x+3) = -4/3

=> -4* ( x+3) = 3

=> -4x -12 =3

=> -4x = 15

=> x = -15/4

**Therefore x = 0 and x = -15/4**