Solve for "x" 2 PROBLEMS. EASY 10 POINTS!?

Question

1. (x +2)^2 + 5= (x+3)^2

2.(x+2)^2-x^2= 4(x+1)

Whoever gets the right answer and shows their work will get ten points. Thanks for answering.

Answer 1

(x+2)^2+5= (x+3)^2

(x+2)*(x+2)            (x+3)*(x+3)

x^2+2x+2x+4         x^2+3x+3x+9

x^2+4x+4 =   x^2+6x+9

1. x=2

2. x=0

Answer 2

1. (x +2)^2 + 5= (x+3)^2

    x^2 + 4x + 4 + 5 = x^2 + 6x + 9

    x^2 + 4x - x^2 - 6x + 5 - 9 = 0

    - 2x - 4 = 0

    - 2x = 4

        x = 4/-2

        x = - 2   ANSWER

2.  (x+2)^2-x^2= 4(x+1)

     x^2 + 4x + 4 - x^2 = 4x + 4

     4x + 4 = 4x + 4

     4x - 4x = 4 - 4

              0 = 0   ANSWER

Hope this helps.

teddy boy

    

Answer 3

1.) (x + 2)^2 + 5 = (x + 3)^2 (FOIL)

x^2 + 4x + 4 + 5 = x^2 + 6x + 9 (combine like terms)

x^2 + 4x + 9 = x^2 + 6x + 9 (subtract x^2 and 9 from both sides)

4x = 6x (subtract 4x from both sides)

0 = 2x (divide both sides by 2)

0 = x

x = 0 <===ANSWER

2.) (x + 2)^2 - x^2 = 4(x + 1) (distributive property and FOIL)

x^2 + 4x + 4 - x^2 = 4x + 4 (combine like terms)

4x + 4 = 4x + 4 (subtract 4x and 4 from both sides)

0 = 0

The variable was canceled out of the equation and the resulting equation is true, as 0 does in fact equal 0. Therefore, x = all real numbers (this equation has infinitely many solutions).