Answer:
[tex]\textsf{A)} \quad 5+3\sqrt{2}[/tex]
Step-by-step explanation:
Given rational expression:
[tex]\dfrac{4+\sqrt{2}}{2 -\sqrt{2}}[/tex]
To rationalise the denominator, multiply both the numerator and denominator by the conjugate of the denominator.
The conjugate of an expression is where we change the sign in the middle of the two terms. Therefore, the conjugate of the denominator of the given expression is:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\implies \dfrac{4+\sqrt{2}}{2 -\sqrt{2}} \cdot \dfrac{2 +\sqrt{2}}{2 +\sqrt{2}}[/tex]
Simplify:
[tex]\implies \dfrac{(4+\sqrt{2})(2 +\sqrt{2})}{(2 -\sqrt{2})(2 +\sqrt{2})}[/tex]
[tex]\implies \dfrac{8+4\sqrt{2}+2\sqrt{2}+\sqrt{2}\sqrt{2}}{4+2\sqrt{2}-2\sqrt{2}-\sqrt{2}\sqrt{2}}[/tex]
[tex]\implies \dfrac{8+6\sqrt{2}+2}{4-2}[/tex]
[tex]\implies \dfrac{10+6\sqrt{2}}{2}[/tex]
Reduce to its simplest form by dividing the numerator and the denominator by 2:
[tex]\implies 5+3\sqrt{2}[/tex]
