Question

Solve the given expression $\dfrac{3+\sqrt{72}}{3-\sqrt{72}}$
[a] $\dfrac{9+2\sqrt{2}}{-7}$
[b] $\dfrac{9+4\sqrt{2}}{-7}$
[c] $\dfrac{-9}{7}$
[d] $1+\dfrac{4\sqrt{2}}{9}$

Answer 1

Hint: Rationalise the denominator and use the algebraic identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Simplify the expression. Recall when rationalising the denominator we multiply both the numerator and denominator of the expression by the radical conjugate of the denominator.

Complete step-by-step answer:
Radical conjugate: If $a-\sqrt{b}$ is any real number, where a and b are rational numbers, then the radical conjugate of $a-\sqrt{b}$ is $a+\sqrt{b}$ and the radical conjugate of $a+\sqrt{b}$ is $a-\sqrt{b}$.
Rationalising the denominator: By rationalising the denominator, we mean to change the given fraction to another equal fraction in which the denominator is rational. When rational the fraction with a denominator of type $a+\sqrt{b}$ and $a-\sqrt{b}$ we multiply both the numerator and denominator of the fraction by the radical conjugate of the denominator and then simplify. Since $\left( a+\sqrt{b} \right)\left( a-\sqrt{b} \right)={{a}^{2}}-b$ and $\left( a-\sqrt{b} \right)\left( a+\sqrt{b} \right)={{a}^{2}}-b$, the denominator is converted into a rational number or in other words, the denominator of the fraction is rationalised.
In the given fraction, we have the denominator $3-\sqrt{72}$. So the radical conjugate is $3+\sqrt{72}$.
Multiplying numerator and denominator of the above fraction by $3+\sqrt{72}$, we have
Denominator $=\left( 3-\sqrt{72} \right)\left( 3+\sqrt{72} \right)=9-72=-63$
Numerator $=\left( 3+\sqrt{72} \right)\left( 3+\sqrt{72} \right)=9+72+6\sqrt{72}=81+6\sqrt{72}$
Now we know that $72={{2}^{3}}{{3}^{2}}$
Hence, we have $\sqrt{72}=6\sqrt{2}$
Hence Numerator $=81+36\sqrt{2}$
Hence the fraction becomes $=\dfrac{81+36\sqrt{2}}{-63}$
Dividing numerator and denominator by 9, we get
$\dfrac{3+\sqrt{72}}{3-\sqrt{72}}=\dfrac{9+4\sqrt{2}}{-7}$
Hence option [b] is correct.
Note: Alternatively we have

$\sqrt{72}=6\sqrt{2}$
Hence $\dfrac{3+\sqrt{72}}{3-\sqrt{72}}=\dfrac{3+6\sqrt{2}}{3-6\sqrt{2}}=\dfrac{1+2\sqrt{2}}{1-2\sqrt{2}}$
Multiply the numerator and the denominator by $1+2\sqrt{2}$ and simplify to get the result.