Question

How do you simplify and find the restrictions for $ \dfrac{{{x^2} + x}}{{{x^2} + 2x}}? $

Answer 1

Hint: As we know that this type of question is based on the concept of polynomial. Here in this question we have to consider the whole function i.e. both the numerator and the denominator. We will first take out the common factor from the numerator and the denominator and then cancel the common term. Then we will simplify the equation and to find the restriction we have to set the denominator which is not equal to zero. And then we solve and find the values of $ x $ for which the denominator is equal to zero which are the restrictions of the function.

Complete step by step solution:
According to the question we have to simplify and find the restrictions of
 $ \dfrac{{{x^2} + x}}{{{x^2} + 2x}} $ .
First we will simplify the given equation, let us now take the $ x $ common from the numerator and the denominator. It gives us
 $ \dfrac{{{x^2} + x}}{{{x^2} + 2x}} = \dfrac{{x(x + 1)}}{{x(x + 2)}} $ .
We will now cancel the common factor and it gives the new equation which is $ \dfrac{{x + 1}}{{x + 2}} $ .
Now we have to find the restriction of the function and we know that a function is not defined when the denominator is equal to zero. So to find the restriction we have to set the denominator to not equal to $ 0 $ .
It means that $ {x^2} + 2x \ne 0 $ . We will now take the common factor from the two terms of the expression i.e. $ x(x + 2) \ne 0 $ .We know that if $ ab \ne 0 $ , then either $ a \ne 0 $ or $ b \ne 0 $ . So by applying this we can write $ x \ne 0 $ and $ x + 2 \ne 0 \Rightarrow x \ne - 2 $ .
Hence the simplified form of the equation is $ \dfrac{{x + 1}}{{x + 2}} $ and the restrictions are $ x \ne - 2,0 $ .

Note: We should note that whenever we get this type of question , we should be careful that we do not have to find the restrictions of the equation from the simplified form, we might not get all the restrictions. Also we have to be careful to avoid any kind of calculation mistakes based on the positive and the negative sign convention.