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Question: How do you simplify \[\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3x}\]?...

How do you simplify x2−9x2+3x\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3x}?

Explanation

Solution

Consider the numerator and use the algebraic identity: - a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to simplify the expression: - x2−9{{x}^{2}}-9. Now, take x common from the terms in the denominator and simplify it. Cancel the common terms of the numerator with the common terms of the denominator to get the simplified form.

Complete answer:
Here, we have been asked to simplify the expression: - x2−9x2+3x\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3x}. Let us assume this expression as E. So, we have,
⇒E=x2−9x2+3x\Rightarrow E=\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3x}
Now, as we can see that we have the expression x2−9{{x}^{2}}-9 in the numerator. Here, 9 can be written as 32{{3}^{2}}, so the expression becomes,
⇒E=x2−32x2+3x\Rightarrow E=\dfrac{{{x}^{2}}-{{3}^{2}}}{{{x}^{2}}+3x}
Clearly, we have the numerator of the form a2−b2{{a}^{2}}-{{b}^{2}} whose factored form is given by the algebraic identity: - a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right). So, using this identity, we get,
⇒E=(x+3)(x−3)x2+3x\Rightarrow E=\dfrac{\left( x+3 \right)\left( x-3 \right)}{{{x}^{2}}+3x}
Now, taking x common from the terms in the denominator, we get,
⇒E=(x+3)(x−3)x(x+3)\Rightarrow E=\dfrac{\left( x+3 \right)\left( x-3 \right)}{x\left( x+3 \right)}
Cancelling the like terms (x+3)\left( x+3 \right) from both the numerator and the denominator, we get,
⇒E=x−3x\Rightarrow E=\dfrac{x-3}{x}
Breaking the terms, we get,
⇒E=xx−3x\Rightarrow E=\dfrac{x}{x}-\dfrac{3}{x}
⇒E=1−3x\Rightarrow E=1-\dfrac{3}{x}
Hence, the simplified form of the given expression is: - x2−9x2+3x=1−3x\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3x}=1-\dfrac{3}{x} and which is our answer.

Note: One may note that whenever we are asked to simplify a given expression which is algebraic in nature then we use basically three formulas of algebraic identities according to the given situation. These identities are: - a2−b2=(a+b)(a−b),a2+b2−2ab=(a−b)2{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right),{{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}, a2+b2+2ab=(a+b)2{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}. In the above question we have used the first formula because the other two identities are not applicable for the given situation. Do not forget to cancel the common factor whenever possible. Sometimes we also use the middle term split method when the above mentioned whole square formulas are not applicable.