Question

rationalize the denominator:
8/3+√x

Answer 1

\[\large \frac{8}{3+\sqrt{x}}\]



\[\large \frac{8}{3+\sqrt{x}}\times\frac{3-\sqrt{x}}{3-\sqrt{x}}\]



\[\large \frac{8(3-\sqrt{x})}{(3+\sqrt{x})(3-\sqrt{x})}\]



\[\large \frac{24-8\sqrt{x}}{9-x}\]

So \[\large \frac{8}{3+\sqrt{x}}=\frac{24-8\sqrt{x}}{9-x}\]

Answer 2

\[8(3-\sqrt(x))/(3+\sqrt(x))(3-\sqrt(x))\]
\[24-8\sqrt(x)/(9-x)\]

Answer 3

when you have a denominator a+b, you multiply both numerator and denominator with a-b. if denominator = a-b, multiply a+b.
this is to remove the radical sign below the bar(denominator).