Chapter 6.1, Problem 16WE

To determine

To calculate: The simplified form of the expression $\frac{{\left(a-5\right)}^2}{25-a^2}$ and state restrictions on variable.

Answer to Problem 16WE

The simplified value of the expression $\frac{{\left(a-5\right)}^2}{25-a^2}$ is $\frac{5-a}{5+a}$ and restriction is $a\ne \pm 5$.

Explanation of Solution

Given information:

The expression $\frac{{\left(a-5\right)}^2}{25-a^2}$.

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

Consider the expression $\frac{{\left(a-5\right)}^2}{25-a^2}$.

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, take out $-1$ common from numerator and factorize denominator using the formula $X^2-Y^2=\left(X-Y\right)\left(X+Y\right)$ and then strike out common terms from the expression and simplify it as,

  $\begin{array}{c}\frac{{\left(a-5\right)}^2}{25-a^2}=\frac{{\left(-1\right)}^2{\left(5-a\right)}^2}{\left(5-a\right)\left(5+a\right)}\\ =\frac{5-a}{5+a}\end{array}$

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\begin{array}{c}25-a^2=0\\ a^2=25\\ a=\pm 5\end{array}$

For denominator not to be zero, $a\ne \pm 5$.

Thus, the simplified value of the expression $\frac{{\left(a-5\right)}^2}{25-a^2}$ is $\frac{5-a}{5+a}$ and restriction is $a\ne \pm 5$.