Question

How do you solve $1=\dfrac{1}{{{x}^{2}}+2x}+\dfrac{x-1}{x}$ and check for extraneous solutions?

Answer 1

Hint: In this question we have an expression which is in the form of a polynomial equation. We will first take the lowest common multiple in the right-hand side of the expression and then simplify the expression and factorize the expression to get the required solutions for the expression. On getting the solution we will also check for any extraneous solutions.

Complete step by step solution:
We have the expression as:
$\Rightarrow 1=\dfrac{1}{{{x}^{2}}+2x}+\dfrac{x-1}{x}$
Since in the right-hand side of the expression we have fractions, we will take the lowest common multiple by cross multiplying the terms as:
$\Rightarrow 1=\dfrac{1\times x+(x-1)({{x}^{2}}+2x)}{x({{x}^{2}}+2x)}$
On multiplying the terms, we get:
$\Rightarrow 1=\dfrac{x+(x\times {{x}^{2}}-1\times {{x}^{2}}+x\times 2x-1\times 2x)}{x\times {{x}^{2}}+x\times 2x}$
On simplifying the terms, we get:
$\Rightarrow 1=\dfrac{x+{{x}^{3}}-{{x}^{2}}+2{{x}^{2}}-2x}{{{x}^{3}}+2{{x}^{2}}}$
On transferring the term ${{x}^{3}}+2{{x}^{2}}$ from the right-hand side to the left-hand side, we get:
$\Rightarrow {{x}^{3}}+2{{x}^{2}}=x+{{x}^{3}}-{{x}^{2}}+2{{x}^{2}}-2x$
On transferring the terms ${{x}^{3}}+2{{x}^{2}}$ from the left-hand side to the right-hand side, we get:
$\Rightarrow 0=x+{{x}^{3}}-{{x}^{3}}-{{x}^{2}}+2{{x}^{2}}-2{{x}^{2}}-2x$
On simplifying the terms, we get:
$\Rightarrow 0=x-{{x}^{2}}-2x$
On subtracting, we get:
$\Rightarrow 0=-{{x}^{2}}-x$
Now transferring the terms on the right-hand side, to the left-hand side, we get:
$\Rightarrow {{x}^{2}}+x=0$
Now in the left-hand side, we have the term $x$ common in both the terms therefore, on taking it out as common, we get:
$\Rightarrow x(x+1)=0$
The above expression is in the factored form therefore, we can say:
$x=0$ or $x+1=0$
Which can be written as:
$x=0$ or $x=-1$

Now we can see that when $x=0$, the denominator of the left-hand side of the expression will become $0$, which is a fallacy therefore the solution $x=0$ is an extraneous solution. This means that the only solution for the expression is $x=-1$, which is the required answer.

Note: It is to be remembered in mathematics that when $ab=0$, either $a=0$ or $b=0$. This rule is implied when factors of an expression equal to $0$. It is to be remembered that the term extraneous solutions are those solutions which do not give us the required values which are to be expected from the equation.