- WELCOME TO OUR VIDEO ON SOLVING RADICAL EQUATIONS.
THIS VIDEO WILL FOCUS ON SOLVING EQUATIONS WITH ONE RADICAL.
THERE'S A SECOND VIDEO THAT DEALS WITH SOLVING
RADICAL EQUATIONS WITH TWO RADICALS.
SO HERE ARE THE STEPS WE'LL FOLLOW
TO SOLVE THIS TYPE OF EQUATION.
NUMBER 1, WE WILL ISOLATE THE RADICAL
ON ONE SIDE OF THE EQUATION.
THEN WE'RE GOING TO RAISE BOTH SIDES OF THE EQUATION
TO THE POWER OF THE INDEX.
THEN WE WILL SOLVE AND THEN CHECK OUR SOLUTIONS.
AND THE REASON IT'S SO IMPORTANT TO CHECK OUR SOLUTIONS THIS TIME
IS SOME SOLUTIONS MAY NOT WORK
AND THESE ARE CALLED EXTRANEOUS SOLUTIONS.
NOW YOU MAY BE ASKING YOURSELF
WHY WOULD WE RAISE BOTH SIDES OF THE EQUATION
TO THE POWER OF THE INDEX.
WELL IF YOU HAVE THE SQUARE ROOT OF X, WHICH HAS AN INDEX OF 2,
AND YOU SQUARE IT, THIS WOULD JUST EQUAL X.
SQUARING AND SQUARE ROOTING ARE OPPOSITE OPERATIONS
SO THESE TWO UNDO EACH OTHER LEAVING US WITH JUST X
WHICH OBVIOUSLY MAKES SOLVING THE EQUATION MUCH EASIER.
THE SAME THING IS TRUE FOR ANY INDEX.
IF WE HAVE THE CUBE ROOT OF X, AND WE CUBE IT,
THIS IS JUST EQUAL TO X.
SO LET'S AHEAD AND TRY A FEW OF THESE EQUATIONS.
NUMBER 1, THE SQUARE ROOT IS ALREADY ISOLATED.
REMEMBER THE SQUARE ROOT HAS AN INDEX OF 2,
SO WHAT WE'LL DO IS WE'LL SQUARE BOTH SIDES OF THE EQUATION
AND THEN SOLVE FOR X AND CHECK.
AGAIN IF WE SQUARE THE SQUARE ROOT OF X - 7
THIS WILL GIVE US X - 7 WHICH MUST EQUAL 11 SQUARED IS 121
ADD 7 TO BOTH SIDES.
WE HAVE X = 128.
WE CAN GO AHEAD AND CHECK THIS.
DOES THE SQUARE ROOT OF 128 - 7 THAT'S 121, EQUAL 11
AND IT DOES, SO IT CHECKS.
ON NUMBER 2, WE NEED TO ADD 7 TO BOTH SIDES
TO ISOLATE THE SQUARE ROOT BEFORE WE SQUARE BOTH SIDES.
SO IF WE ADD 7 TO BOTH SIDES,
WE WOULD HAVE THE SQUARE ROOT OF 3X + 2 = +7.
I ALWAYS LIKE TO PUT THE INDEX IN HERE
JUST TO REMIND ME THAT I'M GOING TO SQUARE BOTH SIDES.
SO THE SQUARE ROOT OF 3X + 2 SQUARED
WILL BE 3X + 2 = 7 SQUARE ROOT, THAT'S 49,
SUBTRACT 2 ON BOTH SIDES, THAT WOULD GIVE US 3X = 47
DIVIDING BY 3 WE HAVE X = 47 THIRDS.
NOW YOU MIGHT THINK THAT'S GOING TO BE HARD TO CHECK
BUT LOOK WHAT HAPPENS WHEN YOU SUB IN 47/3
THEN YOU MULTIPLY BY 3.
THOSE 3'S ARE GOING TO SIMPLIFY OUT SO THIS WOULD BECOME 47 + 2
WHICH IS 49,
THE SQUARE ROOT OF 49 WOULD BE 7 - 7 = 0
SO THIS CHECKS AS WELL.
LET'S GO AHEAD AND TRY A COUPLE MORE.
AGAIN THE FIRST STEP HERE IS TO ISOLATE THE RADICAL
OR IN THIS CASE, THE CUBE ROOT.
SO WE SHOULD ADD 8 TO BOTH SIDES.
NEXT THIS 2 IS ATTACHED BY MULTIPLICATION
SO WE SHOULD DIVIDE BOTH SIDES OF THE EQUATION BY 2.
SO WE HAVE THE CUBE ROOT 5X - 1 MUST EQUAL 4.
NOW OUR INDEX IS 3
SO WE'LL RAISE BOTH SIDES OF THE EQUATION TO THE POWER OF 3.
THE CUBE ROOT OF 5X - 1 CUBED
WILL JUST BE 5X - 1 = 4 TO THE THIRD POWER WILL BE 64,
SO WE'LL ADD 1 TO BOTH SIDES.
THAT WOULD BECOME 65 DIVIDED BY 5 X EQUALS--
65 DIVIDED BY 5 WOULD BE 13.
AND WE CAN GO AHEAD AND CHECK THIS AS WELL.
2 x THE SQUARE ROOT OF 5 x 13 WOULD BE 65 - 1
THAT WOULD BE 64 - 8 = 0.
AND AGAIN THIS IS THE CUBE ROOT.
SO THE CUBE ROOT OF 64 WOULD BE 4
SO 2 x 4 IS 8 - 8 IS 0, CHECKS.
AND LET'S TAKE A LOOK AT ONE MORE.
REMEMBER BEFORE WE START WE MUST ISOLATE THE SQUARE ROOTS
SO WE'RE GOING TO SUBTRACT 6 ON BOTH SIDES.
SO WE'LL HAVE THE SQUARE ROOT OF X = X - 6.
THE SQUARE ROOTS WILL SQUARE BOTH SIDES OF THE EQUATION.
WE'RE NOT SQUARING EACH TERM, WE'RE SQUARING BOTH SIDES.
SO ON THE LEFT WE'RE GOING TO HAVE X, THAT'S GOOD.
ON THE RIGHT WE ACTUALLY HAVE TO MULTIPLY THIS OUT X - 6 x X - 6.
REMEMBER YOU CAN'T JUST SQUARE X AND SQUARE 6.
SO WE HAVE X = THIS WILL BE X SQUARED
MINUS 6X - 6X IS - 12X + 36.
OKAY WE HAVE A QUADRATIC EQUATION NOW
SO WE'RE GOING TO SET IT EQUAL TO 0
AND SEE IF WE CAN FACTOR IT.
SO IF WE SUBTRACT X ON BOTH SIDES,
WE'RE GOING TO HAVE 0 = X SQUARED - 13X + 36
AND BELIEVE THIS IS FACTORABLE.
FACTORS OF X SQUARED, X, X OTHER FACTORS OF 36 THAT ADD TO -13
AND THERE ARE -4 AND -9.
OKAY, SO IT LOOKS LIKE WE MAY HAVE TWO SOLUTIONS
EITHER X = 4, OR X = 9
BUT LET'S GO AHEAD AND CHECK THESE TO MAKE SURE.
LET'S CHECK X = 4 FIRST.
WE'D HAVE THE SQUARE ROOT OF 4 + 6 = 4.
WELL THIS DOESN'T WORK, THAT WOULD BE 2 + 6 = 4,
THAT'S NOT TRUE
WHICH MEANS THIS IS ONE OF THOSE EXTRANEOUS SOLUTIONS
SO WE CANNOT INCLUDE THAT AS A SOLUTION.
LET'S GO AHEAD AND CHECK X = 9
SO WE'D HAVE THE SQUARE ROOT OF 9 + 6 = 9
AND THIS LOOKS PRETTY GOOD.
THIS WOULD BE 3 + 6 = 9 AND THAT CHECKS.
SO WE DO HAVE ONE SOLUTION
BUT THIS IS A PRIME EXAMPLE
OF WHY IT'S SO IMPORTANT THAT WE DO CHECK OUR ANSWERS.
OKAY, THE NEXT VIDEO WILL DEAL WITH SOLVING
THESE TYPES OF EQUATIONS
WHEN YOU HAVE 2 RADICALS INSTEAD OF JUST 1.
THANK YOU FOR WATCHING, I HOPE YOU FOUND IT HELPFUL.