# Solve Each of the Following Equation and Also Verify Your Solution: (X + 2)(X + 3) + (X − 3)(X − 2) − 2x(X + 1) = 0 - Mathematics

Sum

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

#### Solution 1

$(x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0$
$\text{ or }x^2 + 5x + 6 + x^2 - 5x + 6 - 2 x^2 - 2x = 0$
$\text{ or }12 - 2x = 0$
$\text{ or }x = \frac{12}{2} = 6$
$\text{ Verification: }$
$\text{ L . H . S . }= (6 + 2)(6 + 3) + (6 - 3)(6 - 2) - 2 \times 6(6 + 1)$
$= 72 + 12 - 84 = 0 =\text{ R . H . S .}$

#### Solution 2

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

⇒ [x2 + (2 + 3)x + 2 × 3] + [x2 + (-3 - 2)x + (-3)(-2)] - 2x2 - 2x ∴

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

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

= - 2x + 12 = 0

Substracting 12 from both sides

-2x + 12 - 12 = 0 - 12 ⇒ - 2x = -12

Dividing by -2

"-2x"/-2 = (-12)/-2 => x = 6

∴ x = 6

Verification:

L.H.S. = (x + 2)(x + 3) + (x - 3)(x - 2) - 2x (x + 1)

= (6 + 2)(6 + 3) + (6 - 3)(6 - 2) - 2 × 6(6 + 1)

= 8 × 9 + 3 × 4 - 12 × 7

= 72 + 12 - 84 = 84 - 84 = 0 = R.H.S.

Is there an error in this question or solution?
Chapter 14: Linear Equations in one Variable - Exercise 14 (C) [Page 169]

#### APPEARS IN

Selina Concise Mathematics Class 8 ICSE
Chapter 14 Linear Equations in one Variable
Exercise 14 (C) | Q 1.03 | Page 169
RD Sharma Class 8 Maths
Chapter 9 Linear Equation in One Variable
Exercise 9.1 | Q 6 | Page 5

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