Question

How do you simplify \[\dfrac{{{x}^{2}}+3x-18}{{{x}^{2}}-36}\]?

Answer 1

Hint: The given two quadratic equations need to be simplified. The equations can be factored into the multiplication form of two linear equations. If there is any common factor then we eliminate that to find the simplified form of the given expression \[\dfrac{{{x}^{2}}+3x-18}{{{x}^{2}}-36}\].

Complete step by step solution:
We have been given the division of two quadratic equations. We need to find the factor of the equations.
The two quadratic equations are \[{{x}^{2}}+3x-18\] and \[{{x}^{2}}-36\].
 We use vanishing methods to solve them.
We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{2}}+3x-18=0$.
We take $x=a=3$. We can see $f\left( 3 \right)={{3}^{2}}+3\times 3-18=9+9-18=0$.
So, the root of the $f\left( x \right)={{x}^{2}}+3x-18$ will be the function $\left( x-3 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x-3 \right)$ is a factor of the polynomial \[{{x}^{2}}+3x-18\].
So, \[{{x}^{2}}+3x-18=\left( x-3 \right)\left( x+6 \right)\].
For $g\left( x \right)={{x}^{2}}-36$, we use the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Putting the values of $a=x;b=6$, we get \[{{x}^{2}}-36=\left( x-6 \right)\left( x+6 \right)\]
Now we put the factored values for the two equations.
We get \[\dfrac{{{x}^{2}}+3x-18}{{{x}^{2}}-36}=\dfrac{\left( x-3 \right)\left( x+6 \right)}{\left( x-6 \right)\left( x+6 \right)}\].
Now we can see in the denominator and the numerator the common factor is $\left( x+6 \right)$.
We can eliminate the factor from both denominator and the numerator.
So, \[\dfrac{{{x}^{2}}+3x-18}{{{x}^{2}}-36}=\dfrac{\left( x-3 \right)\left( x+6 \right)}{\left( x-6 \right)\left( x+6 \right)}=\dfrac{\left( x-3 \right)}{\left( x-6 \right)}\]. The only condition being $x\ne -6$.

Therefore, the simplified form of \[\dfrac{{{x}^{2}}+3x-18}{{{x}^{2}}-36}\] is \[\dfrac{\left( x-3 \right)}{\left( x-6 \right)}\].

Note: We can verify the factors for the functions.
We can validate the other solution of $x=-6$ for $f\left( x \right)={{x}^{2}}+3x-18$. We put the value of $x=-6$ in the equation.
$f\left( -6 \right)={{\left( -6 \right)}^{2}}+3\left( -6 \right)-18=36-18-18=0$.
Similarly, the root value $x=\pm 6$ satisfies $g\left( x \right)={{x}^{2}}-36$.