Question

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

Answer 1

Hint: In order to simplify the expression , take out x common from both numerator and denominator and find the restriction by putting denominator not equal to zero.

Complete step-by-step answer:
We are given a mathematical expression one variable $ x $ in the term.
Let’s suppose the function given be $ f\left( x \right) $
 $ f\left( x \right) = \dfrac{{{x^2} + x}}{{{x^2} + 2x}} $
First we have Simplify the expression, by taking common $ x $ from both numerator and denominator
 $ f\left( x \right) = \dfrac{{x(x + 1)}}{{x(x + 2)}} $
Cancelling out $ x $ from both numerator and denominator
 $
  f\left( x \right) = \dfrac{{{x}(x + 1)}}{{{x}(x + 2)}} \\
   = \dfrac{{(x + 1)}}{{(x + 2)}} \;
  $
Since, this expression is in the form of rational number and rational number have a condition that the denominator should not be equal to 0.
The restriction in our expression consist of value of $ x $ which make up the denominator equal to zero
Setting up the expression’s denominator equal to 0 to find the restriction
 $
  {x^2} + 2x \ne 0 \\
  x(x + 2) \ne 0 \\
  x \ne 0\,and\,x + 2 \ne 0 \Rightarrow x \ne - 2 \;
  $
Therefore the simplification of the expression is $ \dfrac{{(x + 1)}}{{(x + 2)}} $ with restriction $ x \ne 0, - 2 $
So, the correct answer is “ $ x \ne 0, - 2 $ ”.

Note: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic. The degree of the quadratic equation is of the order 2.