- WELCOME TO A LESSON ON SIMPLIFYING RADICAL EXPRESSIONS.
FIRST, WE'LL TAKE A LOOK AT SIMPLIFYING RADICALS
THAT DON'T INVOLVE FRACTIONS.
A RADICAL IS NOT SIMPLIFIED
IF THE RADICAND OR THE NUMBER UNDERNEATH THE RADICAL
CONTAINS PERFECT nth ROOT FACTORS.
SO WHAT THAT MEANS IS IF WE HAVE A SQUARE ROOT
WE CAN'T HAVE ANY PERFECT SQUARE FACTORS UNDER THE RADICAL
AND IF WE HAVE A CUBE ROOT
WE CAN'T HAVE ANY PERFECT CUBE FACTORS UNDER THE RADICAL
IF IT'S IN SIMPLIFIED FORM.
SO HERE'S HOW WE'LL ACCOMPLISH SIMPLIFYING RADICALS.
WE'RE GOING TO WRITE THE PRIME FACTORIZATION OF THE RADICAND
AND THEN WE'LL CIRCLE THE nth ROOT FACTORS
WHICH MEANS WE'LL CIRCLE THE EQUAL FACTORS IN GROUPS
THE SAME NUMBER AS THE INDEX.
THE CIRCLED FACTORS ARE THE PERFECT nth ROOTS
THAT CAN BE SIMPLIFIED,
AND THE FACTORS NOT CIRCLED WILL REMAIN UNDER THE RADICAL,
AND THEN ONCE WE DO THIS WE'LL RE-WRITE THE RADICAL
IN SIMPLIFIED FORM.
NOW ONE THING WE DO HAVE TO REMEMBER,
IF THE INDEX IS EVEN
AND THE RADICAND HAS VARIABLES THAT SIMPLIFY
WE DO NEED TO INCLUDE AN ABSOLUTE VALUE
AROUND THE SIMPLIFIED VARIABLES WITH ODD EXPONENTS
TO ASSURE THE VARIABLE IS NON-NEGATIVE.
LET'S TAKE A LOOK AT SOME EXAMPLES.
IF WE WANT TO SIMPLIFY THE SQUARE ROOT OF 50,
REMEMBER THE INDEX WOULD BE 2,
WE'RE GOING TO WRITE THIS OUT IN PRIME FACTOR FORM.
50 IS 2 x 25 AND 25 IS 5 x 5.
SINCE THE INDEX IS 2,
WE'LL CIRCLE THE EQUAL FACTORS IN GROUPS OF 2 AS WE SEE HERE.
THIS IS A PERFECT SQUARE FACTOR OF 50
AND THEREFORE, THIS WILL SIMPLIFY.
THE SQUARE ROOT OF 25 WOULD BE 5
AND WE'RE LEFT WITH THE SQUARE ROOT OF 2.
SO WHAT YOU'LL NOTICE IS EVERYTIME WE HAVE A GROUP
OF EQUAL FACTORS
IT'LL SIMPLIFY PERFECTLY
TO JUST ONE OF THE FACTORS OUTSIDE THE RADICAL.
NUMBER 2, WE HAVE THE CUBE ROOT OF 64,
SO WE'LL WRITE OUT THE PRIME FACTORIZATION OF 64.
I KNOW 64 IS 8 x 8 AND 8 IS (2 x 2) x 2.
SO THERE'S 8, THERE'S ANOTHER 8, SO THAT WOULD BE 64.
AGAIN, THE INDEX IS NOW 3, SO WE'LL CIRCLE GROUPS OF 3.
THERE'S A PERFECT CUBE FACTOR
AND THERE'S A PERFECT CUBE FACTOR.
WELL THE CUBE ROOT OF 2 CUBED WOULD BE 2 FROM THIS GROUP
AND THEN THE CUBE ROOT OF 2 CUBED HERE
WOULD ALSO BE ANOTHER 2.
WELL, 2 x 2 = 4, THE CUBE ROOT OF 64 = 4.
AGAIN, NOTICE WE'RE TAKING ONE FACTOR FROM EACH GROUP
THAT WE CIRCLE.
OKAY, LET'S GO AHEAD AND TRY TWO MORE.
NOW WE HAVE SOME VARIABLES
SO WE DO HAVE TO BE AWARE WE MAY NEED AN ABSOLUTE VALUE
ON THE PART THAT SIMPLIFIES.
LET'S GO AHEAD AND WRITE OUT THE PRIME FACTORIZATION.
8 = (2 x 2) x 2, WE HAVE 4 FACTORS OF X,
7 FACTORS OF Y.
THIS IS A LOT OF WRITING BUT I THINK IT IS WORTH IT
WHEN WE'RE FIRST LEARNING HOW TO DO THIS.
BECAUSE THE INDEX IS 2, WE'LL CIRCLE GROUPS OF 2.
REMEMBER THE CIRCLED FACTORS
REPRESENT THE PERFECT SQUARE FACTORS OF THIS RADICAND.
SO WHEN WE SIMPLIFY THIS, THIS WELL BE 2,
WE'LL HAVE A FACTOR OF X FROM THIS GROUP,
ANOTHER FACTOR OF X FROM THIS GROUP.
THAT WOULD BE X/SQUARED.
NOTICE THAT WE HAVE A FACTOR OF Y FROM HERE, HERE, AND HERE.
THAT WOULD BE Y TO THE 3rd
AND THEN WE HAVE THE SQUARE ROOT OF 2Y.
NOTICE THE 2 AND THE Y WERE NOT CIRCLED.
NOW BECAUSE THE INDEX IS EVEN AND THIS HAS AN ODD EXPONENT
WE DO NEED TO TAKE THE ABSOLUTE VALUE OF THIS.
SO WE COULD TAKE THE ABSOLUTE VALUE OF Y/CUBED
OR WE COULD TAKE THE ABSOLUTE VALUE OF 2X/SQUARED Y/CUBED
IT REALLY DEPENDS ON THE TEXTBOOK.
OKAY, LET'S GO AHEAD AND TRY NUMBER 4.
64 IS 6 x 9, 6 IS 2 x 3, 9 IS 3 x 3.
LOOK AT THE VARIABLES THIS TIME.
WE KNOW WE WANT GROUPS OF 2,
SO INSTEAD OF WRITING 3 FACTORS OF "A"
I'M GOING TO WRITE A/SQUARED.
THERE'S A GROUP OF 2 FACTORS OF A x A.
AND INSTEAD OF WRITING OUT B TO THE 10th,
I'LL WRITE OUT B TO THE 2nd 5 TIMES.
THIS SAVES US SOME WRITING AS LONG AS YOU CAN REMEMBER
THAT EACH OF THESE IS A GROUP OF 2.
SO THERE'S A GROUP OF 2, THERE'S A GROUP OF 2.
THERE'S 5 GROUPS OF 2 FOR THE FACTORS OF B.
SO WHAT DOES THIS SIMPLIFY TO?
WE HAVE A FACTOR OF 3, A, B TO THE 5th
AND WE'RE LEFT WITH THE PART THAT DOESN'T SIMPLIFY
WHICH IS 2A UNDERNEATH THE SQUARE ROOT.
WE HAVE AN EVEN INDEX AND SOME ODD EXPONENTS.
SO WE DO HAVE TO HAVE THE ABSOLUTE VALUE
OF 3AB TO THE 5th.
OKAY, LET'S TAKE A LOOK AT ALL TYPES OF RADICAL EXPRESSIONS.
ALL OF THESE STATEMENTS MUST BE TRUE
IN ORDER FOR A RADICAL TO BE IN SIMPLIFIED FORM.
ALL POSSIBLE nth POWERED FACTORS HAVE BEEN REMOVED.
WE WERE JUST WORKING ON THAT.
NUMBER 2, NO RADICAL CONTAINS A FRACTION,
AND THEN NUMBER 3, NO DENOMINATOR OF A FRACTION
CONTAINS A RADICAL.
LET'S TAKE A LOOK AT SOME EXAMPLES OF THIS.
FOR NUMBER 1, WE'RE NOT ALLOWED TO HAVE THIS FRACTION
UNDERNEATH THIS SQUARE ROOT.
WELL WE CAN RE-WRITE THIS AS THE SQUARED OF 9
OVER THE SQUARED OF 25
BUT NOW WE'RE NOT ALLOWED TO HAVE THE SQUARE ROOT OF 25
IN THE DENOMINATOR.
BUT LUCKILY THE SQUARE ROOT OF 25 SIMPLIFIES PERFECTLY TO 5
AND OUR NUMERATOR WOULD BE 3.
SO WE'RE OKAY ON THIS ONE.
BUT NUMBER TWO'S A LITTLE BIT DIFFERENT.
IF I RE-WRITE THIS AS THE SQUARE ROOT OF 4
OVER THE SQUARE ROOT OF 5
WE CAN'T HAVE THE SQUARE ROOT OF 5 IN OUR DENOMINATOR
IN SIMPLIFIED FORM.
SO WE HAVE TO THINK TO OURSELVES,
WHAT CAN WE DO TO THIS SO THAT IT SIMPLIFIES PERFECTLY.
WELL IF WE HAD 2 FACTORS OF 5, IT WOULD SIMPLIFY PERFECTLY.
SO WHAT WE CAN DO IS MULTIPLY BY THE SQUARE ROOT OF 5
IN THE DENOMINATOR
AS LONG AS WE DO THE SAME TO THE NUMERATOR.
THE REASON WE'D WANT TO DO THAT IS BECAUSE OUR DENOMINATOR
IS NOW THE SQUARED OF 5 x 5
AND OUR NUMERATOR IS THE SQUARED OF 4 x 5.
THIS IS GREAT BECAUSE NOW OUR DENOMINATOR SIMPLIFIES TO 5,
WELL THE SQUARE ROOT OF 4 WOULD BE 2,
SO WE WOULD HAVE 2 SQUARE ROOT 5.
LET'S LOOK AT ANOTHER EXAMPLE.
NOW WE HAVE THE CUBE ROOT OF 2/3.
THIS IS EQUAL TO THE CUBE ROOT OF 2 OVER THE CUBE ROOT OF 3.
NOW IN ORDER FOR THE DENOMINATOR TO SIMPLIFY PERFECTLY
WE'RE GOING TO NEED 3 FACTORS OF 3.
SO WHAT WE CAN GO AHEAD AND MULTIPLY
BY THE CUBE ROOT OF 3 x 3
THAT WILL GIVE US 3 FACTORS OF 3 WHICH MATCHES THE INDEX.
SO I'M GOING TO DO THE SAME IN THE NUMERATOR.
SO NOW WE WOULD HAVE THE CUBE ROOT OF 3 FACTORS OF 3,
WHICH IS PERFECT
OVER THE CUBE ROOT OF (2 x 3) x 3.
AGAIN, WE'RE ONLY CONCERNED ABOUT HAVING THE RADICAL
IN THE DENOMINATOR.
WE DON'T CARE IF IT'S IN A NUMERATOR.
SO HERE'S OUR GROUP OF 3,
SO OUR DENOMINATOR EQUALS 3
AND OUR NUMERATOR WOULD BE THE CUBE ROOT OF 18.
LET'S TRY ONE MORE.
THIS IS ABOUT AS CHALLENGING AS IT'LL GET
UNTIL WE INTRODUCE THE IDEA OF CONJUGATES.
SO THIS IS EQUAL TO THE CUBE ROOT OF 7.
I WANT TO LOOK AT THIS IN FACTORED FORM
SO I'LL WRITE THIS AS (2 x 2) x (X x Y) x Y.
REMEMBER SINCE THE INDEX IS 3
WE WANT 3 FACTORS OF 2X AND Y IN THE DENOMINATOR.
SO WE NEED ONE MORE FACTOR OF 2, TWO MORE FACTORS OF X,
AND ONE MORE FACTOR OF Y.
OF COURSE THE SAME IN THE NUMERATOR.
SO WHAT WOULD THIS LEAVE US WITH?
IN THE NUMERATOR WE'RE GOING TO HAVE (7 x 2) x (X x X) x Y.
OUR DENOMINATOR NOW IS GOING TO HAVE 3 FACTORS OF 2,
3 FACTORS OF X,
AND 3 FACTORS OF Y
AND THAT'S EXACTLY WHY WE MULTIPLIED BY THIS.
YOU CAN SEE 3, 3, AND 3.
SO THIS SIMPLIFIES PERFECTLY.
OUR DENOMINATOR IS GOING TO BE 2XY
AND OUR NUMERATOR WILL BE THE CUBE ROOT OF 14 X/SQUARED Y.
NOW EVEN THOUGH THESE EXPONENTS ARE ODD
WE DON'T NEED ABSOLUTE VALUE AROUND THIS
BECAUSE THE INDEX WAS ODD TO BEGIN WITH.
IT'S ONLY WHEN THE INDEX IS EVEN THAT WE HAVE THAT SPECIAL RULE.
OKAY, I HOPE, YOU FOUND THIS VIDEO HELPFUL.
THANK YOU AND HAVE A GOOD DAY.