Question

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

Answer 1

Hint: In this question we have been given with a polynomial expression which has its terms factored out. We will simplify the given expression by multiplying the terms and getting individual terms. we will then see the terms which are similar and cancel the terms which have opposite signs. We will then simplify the expression and solve for the value to get the value of $x$.

Complete step by step solution:
We have the expression given to us as:
\[\Rightarrow \left( x+2 \right)\left( x+3 \right)+\left( x-3 \right)\left( x-2 \right)-2x\left( x+1 \right)=0\]
On multiplying the terms in the expression, we get:
$\Rightarrow {{x}^{2}}+2x+3x+6+{{x}^{2}}-3x-2x+6-2{{x}^{2}}-2x=0$
Now on writing the similar terms together, we get:
$\Rightarrow 2{{x}^{2}}-2{{x}^{2}}+5x-5x-2x+12=0$
Now since the same term with opposite signs cancel each other, we get:
$\Rightarrow -2x+12=0$
On transferring the term $12$ from the left-hand side of the expression to the right-hand side, we get:
$\Rightarrow -2x=-12$
On multiplying both the sides of the expression with $-1$, we get:
$\Rightarrow 2x=12$
On transferring the term $2$ from the left-hand side to the right-hand side, we get:
$\Rightarrow x=\dfrac{12}{6}$
On simplifying, we get:
$\Rightarrow x=6$, which is the required solution.

Note: Now to check whether the solution is correct, we will substitute $x=6$ in the equation.
On substituting, we get:
\[\Rightarrow \left( 6+2 \right)\left( 6+3 \right)+\left( 6-3 \right)\left( 6-2 \right)-2\left( 6 \right)\left( 6+1 \right)=0\]
On adding the terms, we get:
\[\Rightarrow \left( 8 \right)\left( 9 \right)+\left( 3 \right)\left( 4 \right)-2\left( 6 \right)\left( 7 \right)=0\]
On multiplying the terms, we get:
\[\Rightarrow 72+12-84=0\]
On simplifying, we get:
$\Rightarrow 0=0$, since the left-hand side is equal to the right-hand side the value $x=6$ is correct.
It is to be remembered that when solving an expression which has multiplication, addition, division and subtraction operations, the rule $PEMDAS$ should be remembered which states the sequence of operations which is parenthesis, exponent, multiplication, division, addition and subtraction. This sequence should be remembered otherwise the answer might go wrong.