Find the value of $x : (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0$

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Solution

Given that,

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

$\Rightarrow x^{2} + 2 x + 3 x + 6 + x^{2} - 3 x - 2 x + 6 - 2 x^{2} - 2 x = 0$

$\Rightarrow 2 x^{2} - 2 x^{2} + 5 x - 5 x - 2 x + 12 = 0$

$\Rightarrow - 2 x + 12 = 0$

$\Rightarrow - 2 x = - 12$

$\Rightarrow x = 6$

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Similar questions

Q. Solve

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

Q. Solve:

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

Solve each of the following equations and also verify your solution :

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

Q. Solve each of the following equation and also verify your solution:
(x + 2)(x + 3) + (x − 3)(x − 2) − 2x(x + 1) = 0

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

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Find the value of x:(x+2)(x+3)+(x−3)(x−2)−2x(x+1)=0

Given that, (x+2)(x+3)+(x−3)(x−2)−2x(x+1)=0 ⇒x2+2x+3x+6+x2−3x−2x+6−2x2−2x=0 ⇒2x2−2x2+5x−5x−2x+12=0 ⇒−2x+12=0 ⇒−2x=−12 ⇒x=6

Find the value of $x : (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0$

Given that,

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

$\Rightarrow x^{2} + 2 x + 3 x + 6 + x^{2} - 3 x - 2 x + 6 - 2 x^{2} - 2 x = 0$

$\Rightarrow 2 x^{2} - 2 x^{2} + 5 x - 5 x - 2 x + 12 = 0$

$\Rightarrow - 2 x + 12 = 0$

$\Rightarrow - 2 x = - 12$

$\Rightarrow x = 6$