Question

Find the value of x in \[(x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0\] .

Answer 1

Hint: We solve this by simply simplifying the given equation. We will get a polynomial of degree two or three. We solve this by factorization method and we will obtain the value of x. Sometimes in this type of problem all the terms will cancel out and we obtain a simple solution. We first simplify the first product terms then solve it.

Complete step-by-step answer:
Given, \[(x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0\] . ----- (1)
If we solve this by directly simplifying some terms that may get missed, it will give us a wrong solution.
So now we simplify terms \[(x + 2)(x + 3)\] , \[(x - 3)(x - 2)\] and \[2x(x + 1)\] separately and substitute this in equation (1).
Now take, \[(x + 2)(x + 3)\]
Expanding the brackets,
 \[ = x(x + 3) + 2(x + 3)\]
Multiplying we get,
 \[ = {x^2} + 3x + 2x + 6\]
Adding the coefficients of x and keep it as one term,
 \[ = {x^2} + 5x + 6\]
Now take, \[(x - 3)(x - 2)\]
Expanding the brackets,
 \[ = x(x - 2) - 3(x - 2)\]
 \[ = {x^2} - 2x - 3x + 6\]
 \[ ={x^2} - (2x + 3x) + 6\]
Adding coefficients of x,
 \[ = {x^2} - 5x + 6\]
Now take, \[2x(x + 1)\]
(Expanding brackets,)
 \[ \Rightarrow 2{x^2} + 2x\] .
Now substituting these simplified terms in the equation (1) we get,
 \[(x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0\]
 \[ \Rightarrow ({x^2} + 5x + 6) + ({x^2} - 5x + 6) - (2{x^2} + 2x) = 0\]
(Negative term multiplied by positive term we get positive term.)
 \[ \Rightarrow {x^2} + 5x + 6 + {x^2} - 5x + 6 - 2{x^2} - 2x = 0\]
Rearranging the terms we get,
 \[ \Rightarrow {x^2} + {x^2} - 2{x^2} + 5x - 5x + 6 + 6 - 2x = 0\]
 \[ \Rightarrow 2{x^2} - 2{x^2} + 5x - 5x + 12 - 2x = 0\]
Cancelling \[2{x^2}\] and \[5x\] , we get:
 \[ \Rightarrow 12 - 2x = 0\]
Rearranging we get,
 \[ \Rightarrow - 2x = - 12\]
Multiply by -1 on both sides,
 \[ \Rightarrow 2x = 12\]
Divide by 2 on both sides, we get:
 \[ \Rightarrow x = \dfrac{{12}}{2}\] (Simple division)
 \[ \Rightarrow x = 6\] .
So, the correct answer is “6”.

Note: While simplifying we all make some mistakes, so remember the BODMAS rule. That is the order of operations, first we solve the terms in the brackets, followed by order (square roots), followed by division, followed by multiplication, followed by addition and lastly subtraction. This is what we did in the above calculation part.