- TO RATIONALIZE THE DENOMINATOR OF A RADICAL EXPRESSION,
WE WANT TO ELIMINATE THE RADICAL FROM THE DENOMINATOR.
AND WHEN THE DENOMINATOR CONTAINS A SUM OR DIFFERENCE
INVOLVING A SQUARE ROOT,
AS WE SEE HERE,
WE'RE GOING TO MULTIPLY BOTH THE NUMERATOR AND DENOMINATOR
BY THE CONJUGATE OF THE DENOMINATOR.
SO FOR OUR FIRST EXAMPLE,
SINCE THE DENOMINATOR IS 3 + SQUARE ROOT 7,
WE'RE GOING TO MULTIPLY BOTH THE NUMERATOR AND DENOMINATOR
BY 3 - SQUARE ROOT OF 7.
AND TO HELP REMIND US HOW TO MULTIPLY,
I THINK IT'S HELPFUL TO PUT THESE IN PARENTHESIS,
AS WELL AS HERE.
SO IN THE NUMERATOR, WE'RE GOING TO DISTRIBUTE -2.
SO I HAVE -2 x 3 THAT'S -6.
AND NOW WE'LL HAVE +2 SQUARE ROOT 7.
AND NOW IN THE DENOMINATOR WE'LL HAVE FOUR PRODUCTS.
FIRST WE'LL HAVE 3 x 3 THAT'S 9.
THEN WE'LL HAVE 3 X -SQUARE ROOT 7
THAT'S -3 SQUARE ROOT 7.
NEXT WE HAVE +SQUARE ROOT 7 x +3 THAT'S +3 SQUARE ROOT 7.
NOTICE HOW THESE TWO TERMS ARE OPPOSITES.
AND THEN WE HAVE +SQUARE ROOT 7 x -SQUARE ROOT 7
THAT'LL BE -THE SQUARE ROOT OF 7 x 7.
WHICH BECAUSE WE HAVE TWO EQUAL FACTORS, WE'LL SIMPLIFY.
SO LOOKING A THE NUMERATOR, WE'LL HAVE -6 + 2 SQUARE ROOT 7.
NOW, THE DENOMINATOR,
NOTICE HOW THE TWO MIDDLE TERMS, AGAIN, ARE OPPOSITES.
SO THIS SIMPLIFIES TO 0.
NOTICE HERE WE HAVE THE SQUARE ROOT OF 7 x 7
OR THE SQUARE ROOT OF 7 SQUARED,
SO THIS SIMPLIFIES TO 7.
SO WE HAVE 9 - 7, WHICH IS 2.
NOW, WE'RE NOT QUITE DONE YET,
BECAUSE THE NUMERATOR AND THE DENOMINATOR
DO SHARE A COMMON FACTOR OF 2.
SO WE'LL GO AHEAD AND FACTOR A 2 OUT OF THE NUMERATOR.
WE'D HAVE 2 X THE QUANTITY -3 + SQUARE ROOT 7, ALL OVER 2.
AND HERE WE HAVE 2/2, WHICH SIMPLIFIED TO 1.
SO FINALLY WE HAVE -3 + SQUARE ROOT 7.
LET'S TAKE A LOOK AT OUR SECOND EXAMPLE.
SO WE'RE CONCERNED ABOUT HAVING THE SQUARE ROOT 3
IN THE DENOMINATOR.
AND SINCE WE HAVE A DIFFERENCE HERE,
WE'LL MULTIPLY BOTH THE NUMERATOR AND DENOMINATOR
BY THE CONJUGATE OF 9 - SQUARE ROOT 3,
WHICH WOULD BE 9 + SQUARE ROOT 3.
AND, AGAIN, WE'LL PUT EVERYTHING IN PARENTHESIS
TO HELP US MULTIPLY CORRECTLY.
SO NOW WE'LL HAVE FOUR PRODUCTS IN THE NUMERATOR
AS WELL AS THE DENOMINATOR.
SO WE'LL HAVE (3 x 9) THAT'S 27.
THEN (3 X SQUARE ROOT 3) THAT'S 3 SQUARE ROOT 3,
(- SQUARE ROOT 2 x 9) THAT'S -9 SQUARE ROOT 2.
AND THEN WE HAVE (-SQUARE ROOT 2 X +SQUARE ROOT 3)
THAT'S -SQUARE ROOT 2 X 3 OR SQUARE ROOT 6.
AND NOW THE DENOMINATOR, WE'LL HAVE (9 X 9) THAT'S 81.
AND THEN (9 X SQUARE ROOT 3) THAT'S +9 SQUARE ROOT 3.
NEXT WE HAVE (-SQUARE ROOT 3 x +9),
WHICH IS -9 SQUARE ROOT 3.
AGAIN, NOTICE HOW WE HAVE OPPOSITES HERE.
AND AT THE LAST PRODUCT WE HAVE
(-SQUARE ROOT 3 x +SQUARE ROOT 3),
WHICH WOULD BE - THE SQUARE ROOT OF 3 x 3,
WHICH WE CAN SEE IS A PERFECT SQUARE,
AND WE'LL SIMPLIFY TO 3.
AND LOOKING AT THE NUMERATOR,
NOTICE HOW THERE ARE NO LIKE RADICANDS
AND NOTHING SIMPLIFIES,
SO NOTHING'S GOING TO CHANGE HERE.
AND THEN FOR THE DENOMINATOR,
AND THESE TWO TERMS ARE OPPOSITES, THAT WOULD BE 0.
SO WE HAVE 81 - THE SQUARE ROOT OF 3 SQUARED OR JUST -3,
WHICH WOULD BE 78.
SO THIS WAY DOES NOT SEEM SIMPLIFIED,
BUT NOTICE HOW THE DENOMINATOR
NO LONGER CONTAINS A SQUARE ROOT.
SO THE DENOMINATOR IS RATIONALIZED.
OKAY, HOPE YOU FOUND THIS HELPFUL.