Question

(iv) (x + 2) (x + 3) + (x - 3) (x - 2) - 2x (x + 1) = 0​

Answer 1

★ Given :

(x+2)(x+3) + (x-3)(x-2) - 2x(x+1) = 0

★ To find:

The value of x

★ Solution :

(x+2)(x+3) + (x-3)(x-2) - 2x(x+1) = 0

{x(x+3)+2(x+3)} + {x(x- 3)-2(x - 3)} - 2x(x+1) = 0

{x² + 3x+2x+6} + {x² - 3x - 2x +6} - 2x² - 2x = 0

{x²+5x+6}+{x² - 5x +6} - 2x² -2x = 0

x² + 5x +6 +x² - 5x +6 -2x² - 2x = 0

(x² + x² -2x²) + (5x-5x-2x) + (6+6) = 0

(2x² - 2x²) -2x + 12 = 0

-2x +12 = 0

2x = 12

$x = \frac{\cancel{12}}{\cancel{2}} \\ \\ \\ {\pink{\boxed{x = 6}}}$

Answer 2

Solution:-

=>(x+2) (x+3) + (x-3) (x-2) -2x(x+1) = 0

=>(x²+3x+2x+6 )+(x²-2x-3x+6)

-2x²-2x = 0

=> x²+5x+6 + x²-5x+6 - 2x² - 2x = 0

we can write it as

=> x²+x²-2x² +5x-5x -2x +6+6 = 0

=> 2x²-2x² - 2x + 12 = 0

=> - 2x + 12 = 0

=> - 2x = - 12

=> 2x = 12

=> x = 12/2

=> x = 6

i hope it helps you.