`(6+2sqrt(7))/(3-sqrt(7))` express in the form of p+q`sqrt(7)` where p and q are integers

We have to express `(6+2sqrt(7))/(3-sqrt(7))` in the form of `p+qsqrt(7)`

let's rationalize the denominator to get rid of the radicals in the expression,

So we have to multiply the numerator and denominator with the conjugate of the denominator ,

Since the denominator is `(3-sqrt(7))` so it's conjugate is `(3+sqrt(7))` ,

`(6+2sqrt(7))/(3-sqrt(7))=((6+2sqrt(7))/(3-sqrt(7)))((3+sqrt(7))/(3+sqrt(7)))`

`=(6(3+sqrt(7))+2sqrt(7)(3+sqrt(7)))/(3^2-(sqrt(7))^2)`

Let's simplify the above expression,

`=(6*3+6*sqrt(7)+2sqrt(7)*3+2sqrt(7)*sqrt(7))/(9-7)`

`=(18+6sqrt(7)+6sqrt(7)+14)/2`

`=(32+12sqrt(7))/2`

`=32/2+(12sqrt(7))/2`

`=16+6sqrt(7)`

So, now the expression is in the form of `(p+qsqrt(7))` where p is 16 and q is 6.

Comments

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