Answer:
[tex]\displaystyle \frac{9}{\sqrt{7}-2}=3\sqrt{7}+6[/tex]
Step-by-step explanation:
Rationalizing the Denominator
Rationalizing the denominator means to get rid of any radicals in the denominator.
We use the conjugate of the denominator to rationalize. The conjugate of a binomial is obtained by changing the sign that is between the two terms while keeping the same order of the terms. For example, the conjugate of
[tex]\sqrt{7}-2[/tex] is [tex]\sqrt{7}+2[/tex].
We need to use the algebraic identity:
[tex](a-b)(a+b)=a^2-b^2[/tex]
Let's rationalize
[tex]\displaystyle \frac{9}{\sqrt{7}-2}[/tex]
Multiply numerator and denominator by the conjugate of the denominator:
[tex]\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9}{\sqrt{7}-2}\cdot\frac{\sqrt{7}+2}{\sqrt{7}+2}[/tex]
Operate in the numerator and apply the identity in the denominator
[tex]\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}[/tex]
[tex]\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9\sqrt{7}+18}{(\sqrt{7})^2-2^2}[/tex]
[tex]\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9\sqrt{7}+18}{7-4}=\frac{9\sqrt{7}+18}{3}[/tex]
Dividing:
[tex]\boxed{\displaystyle \frac{9}{\sqrt{7}-2}=3\sqrt{7}+6}[/tex]
