Question

How do you simplify $\dfrac{x+5}{{{x}^{2}}-25}$ ?

Answer 1

Hint: We have been given a fractional term whose numerator has a linear equation in x-variable and denominator has a quadratic equation in variable-x. In order to simplify this expression, we must focus upon simplifying and breaking the quadratic expression to linear equations and cancel a few repeated terms by applying the basic algebraic properties.

Complete step by step solution:
Given that $\dfrac{x+5}{{{x}^{2}}-25}$.
We know that $25=5\times 5$. Thus, 25 can also be written as the square of 5, that is, $25={{5}^{2}}$.
Using this substitution, we can modify the given equation as
$\Rightarrow \dfrac{x+5}{{{x}^{2}}-25}=\dfrac{x+5}{{{x}^{2}}-{{5}^{2}}}$
We observe that an algebraic property can be used which says that the difference of squares of two terms can be expressed as the product of their sum and difference, that is, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Here, $a=x$ and $b=5$
$\Rightarrow {{x}^{2}}-{{5}^{2}}=\left( x+5 \right)\left( x-5 \right)$
Substituting this value of ${{x}^{2}}-{{5}^{2}}$ in the main equation, we get
 $\Rightarrow \dfrac{x+5}{{{x}^{2}}-25}=\dfrac{x+5}{\left( x+5 \right)\left( x-5 \right)}$
On the right hand side of the equation, both the numerator as well as the denominator consists of the term $x+5$, thus, we will divide and cancel this term from the right hand side completely.
$\Rightarrow \dfrac{x+5}{{{x}^{2}}-25}=\dfrac{1}{\left( x-5 \right)}$
Thus, we are left with only $\left( x-5 \right)$ in the denominator which cannot be simplified further.

Therefore, $\dfrac{x+5}{{{x}^{2}}-25}$ can be simplified as $\dfrac{1}{x-5}$.

Note: In order to solve such mathematical expressions or even complex mathematical expressions, we must have prior knowledge of the basic algebraic identities. Some of the other algebraic properties which are most commonly in use are that of the square of the sum of two terms and the square of the difference of two terms which is ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ respectively.