Question

i am stuck on this question can someone plz help me?
rationalize the denominator of 5/(6+3sqrt2)-(2+2sqrt2)
A.20-5sqrt2/14
B.20+5sqrt2/18
C.20-sqrt2/14
D.5sqrt2/2+4sqrt2

Answer 1

@skullpatrol

Answer 2

@damien

Answer 3

(6+3sqrt2)-(2+2sqrt2) the denominator?

Answer 4

yes

Answer 5

Simplify the denominator first by collecting the like terms.. (sorry I was late, I forgot about this question)

Answer 6

$$\bf
\frac{5}{(6+3\sqrt{2})-(2+2\sqrt{2})}\\
\color{blue}{\text{multiplying both sides by the denominator's conjugate}}\\
\frac{5}{(6+3\sqrt{2})-(2+2\sqrt{2})} \times \frac{(6+3\sqrt{2})+(2+2\sqrt{2})}{(6+3\sqrt{2})+(2+2\sqrt{2})}\\
\color{blue}{\text{keep in mind that } \implies (a-b)(a+b) = (a^2-b^2)}\\
\frac{5\times [(6+3\sqrt{2})+(2+2\sqrt{2})]}{(6+3\sqrt{2})^2-(2+2\sqrt{2})^2}
$$

Answer 7

\[\frac{ 5 }{ \left( 6+3\sqrt{2} \right)-\left( 2+\sqrt{2} \right) }\]
\[\frac{ 5 }{6+3\sqrt{2}-2-2\sqrt{2} }=\frac{ 5 }{4+1\sqrt{2} }\]
\[\frac{ 5 }{ 4+\sqrt{2}} \times \frac{ 4-\sqrt{2} }{4-\sqrt{2}}=\frac{ 5\left( 4-\sqrt{2} \right) }{ 4^{2}-\left( \sqrt{2 }\right)^2 }\]
\[=\frac{ 5\left( 4-\sqrt{2} \right) }{16-2 }=\frac{ 20-5\sqrt{2} }{ 14 }\]

Answer 8

thanks so much for your help i understand better now