Question

Another Algebra Question (in Replies)

Answer 1

Show that \[\frac{ 1 }{ a + b \sqrt{c} } = \frac{ a -b \sqrt{c} }{ a^2 - b^2c } \]
Express the following
\[ \frac{ 2+3 \sqrt{2} } { 9-7 \sqrt{2}} \] as \[x+y \sqrt{n}\] Where x and y are rational and n is integer.

Show Work.
I know the solution But I want to make sure its correct.

Answer 2

The following is from an earlier question which I had validly solved.
\[ \frac{ 1 } { 7 + 5\sqrt{2}} = \frac{ 1 * ( 7 - 5\sqrt{2}) } { (7 + 5\sqrt{2}) ( 7 - 5\sqrt{2})} \]
\[\frac{ 1 * ( 7 - 5\sqrt{2}) } { (7 + 5\sqrt{2}) ( 7 - 5\sqrt{2})} = \frac{7-5\sqrt{2}}{7^2+ ( 35\sqrt{2}-35\sqrt{2}) - 5^2*2} \] Thus \[\frac{ 1 }{ a + b \sqrt{c} } = \frac{ a -b \sqrt{c} }{ a^2 - b^2c } \] Is Valid. Then from there

\[ \frac{ 7 -5 \sqrt{2} }{ 7^2 - 5^22 } = \frac{ 7 -5 \sqrt{2} }{ 49-50 } = \frac{ 7 -5 \sqrt{2} }{-1 } = -7+5\sqrt{2} \]

Answer 3

....so which one of these is the question again?

Answer 4

The first one.
the second one is the form of work which I had done earlier.

Answer 5

yes it looks fine

Answer 6

Verified the problem and doubled checked it using the same method. thanks for the confirmation.