Answer :
[tex](x+1)^2 -(x-1)^2 = 32 \\ \\x^2+2x+1 -(x^2-2x+1)=32\\ \\x^2+2x+1-x^2+2x-1=32\\ \\ 4x =32\ \ / :4 \\ \\x=\frac{32}{4}\\ \\x=8 [/tex]
All the busy-work is on the left side.
1). Take care of the (x+1)² . . .
(x + 1)² = x² + 2x + 1
2). Take care of the (x-1)² . . .
(x - 1)² = x² - 2x + 1
The problem says that (the first expression) minus (the second expression) is 32 .
(x² + 2x + 1) - (x² - 2x + 1) = 32
Can you handle this ?
Only one more touchy point to take care of . . . removing the second parentheses.
Since there's a negative sign before it, that changes the sign of everything inside:
(x² + 2x + 1) PLUS (- x² + 2x - 1) = 32
Now you can just erase all the parentheses:
x² + 2x + 1 - x² + 2x - 1 = 32
Now I'm just going to write the things on the left side in a different order,
without changing anything:
x² - x² + 2x + 2x + 1 - 1 = 32
Isn't that neat ? ! ?
What you have now is:
4x = 32
Divide each side by 4 :
x = 8
Just watch how slick that works:
(x + 1)² - (x - 1)² = ?
(x + 1) = 9 (x - 1) = 7
(9)² - (7)² = 81 - 49 = . . . . . . . . 32 !
спасибо
Удачи
1). Take care of the (x+1)² . . .
(x + 1)² = x² + 2x + 1
2). Take care of the (x-1)² . . .
(x - 1)² = x² - 2x + 1
The problem says that (the first expression) minus (the second expression) is 32 .
(x² + 2x + 1) - (x² - 2x + 1) = 32
Can you handle this ?
Only one more touchy point to take care of . . . removing the second parentheses.
Since there's a negative sign before it, that changes the sign of everything inside:
(x² + 2x + 1) PLUS (- x² + 2x - 1) = 32
Now you can just erase all the parentheses:
x² + 2x + 1 - x² + 2x - 1 = 32
Now I'm just going to write the things on the left side in a different order,
without changing anything:
x² - x² + 2x + 2x + 1 - 1 = 32
Isn't that neat ? ! ?
What you have now is:
4x = 32
Divide each side by 4 :
x = 8
Just watch how slick that works:
(x + 1)² - (x - 1)² = ?
(x + 1) = 9 (x - 1) = 7
(9)² - (7)² = 81 - 49 = . . . . . . . . 32 !
спасибо
Удачи