Question

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Solve (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0


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Answer 1

Given:

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

What to do:

• We need to solve this equation.

Solution:

❐ Here in this question we are given with a equation and we need to solve that equation. let's start our solution and understand the steps to get our final result. Let's do it!

⠀

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

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

→ x²+3x + 2(x + 3)+(x-3) (x - 2) - 2x(x + 1) = 0

→ x²+3x + 2x + 6 + (x-3) (x-2) - 2x(x + 1) = 0

→ x² + 5x + 6 + (x - 3) (x – 2) - 2x(x + 1) = 0

→ x² + 5x + 6 + x² - 5x + 6 - 2x(x + 1) = 0

→ x² + 5x + 6 + x² - 5x + 6 - 2x² - 2x = 0

→ x² + 6 + x² + 6 - 2x² - 2x = 0

→ 0x² + 6 + 6 - 2x = 0

→ 0x² + 12 - 2x = 0

→ 0 + 12 - 2x = 0

→ 12 - 2x = 0

→ -2x + 12 = 0

→ -2x = 0 - 12

→ -2x = -12

→ 2x = 12

→ x = 6.

Hence,

• The required value of x is 6.

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Answer 2

Answer:

Given :-

$\leadsto \sf\bold{\pink{(x + 2) (x + 3) + (x - 3) (x - 2) - 2x(x + 1) =\: 0}}$

To Find :-

• What is the value of x.

Solution :-

$\implies \sf (x + 2) (x + 3) + (x - 3) (x - 2) - 2x(x + 1) =\: 0$

$\implies \sf x(x + 3) + 2(x + 3) + x(x - 2) - 3(x - 2) - 2x(x + 1) =\: 0$

$\implies \sf {x}^{2} + 3x + 2x + 6 + {x}^{2} - 2x - 3x + 6 - 2{x}^{2} - 2x =\: 0$

$\implies \sf {x}^{2} + {x}^{2} - 2{x}^{2} {\cancel{+ 3x}} {\cancel{+ 2x}} {\cancel{- 2x}} {\cancel{- 3x}} - 2x + 6 + 6 =\: 0$

$\implies \sf {\cancel{2{x}^{2}}} {\cancel{- 2{x}^{2}}} - 2x + 12 =\: 0$

$\implies \sf {\cancel{-}} 2x =\: {\cancel{-}} 12$

$\implies \sf 2x =\: 12$

$\implies \sf x =\: \dfrac{\cancel{12}}{\cancel{2}}$

$\implies \sf\bold{\red{x =\: 6}}$

$\therefore$ ${\large{\bold{\purple{\underline{The\: value\: of\: x\: is\: 6.}}}}}$