- TO SOLVE AN EQUATION
CONTAINING ONE RADICAL, 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. THIS WILL UNDO THE RADICAL. THEN WE'LL SOLVE THE RESULT
OF THE EQUATION AND CHECK OUR SOLUTIONS. SO FOR THE FIRST EXAMPLE WE HAVE
THE SQUARE ROOT OF THE QUANTITY X + 6 = X + 4. NOTICE HOW THE SQUARE ROOT
IS ISOLATED ON THE LEFT SIDE
OF THE EQUATION. AND BECAUSE WE HAVE
A SQUARE ROOT, THE INDEX IS 2. SO WE'RE GOING TO SQUARE
BOTH SIDES OF THE EQUATION. NOTICE ON THE RIGHT
WE'RE SQUARING THE ENTIRE SIDE, WE'RE NOT SQUARING EACH TERM. SO WHEN WE SQUARE
THE SQUARE ROOT OF X +6, THIS WILL SIMPLIFY
TO THE RADICAND OF X + 6. AND THEN FOR THE RIGHT SIDE
WE HAVE THE QUANTITY X + 4 x THE QUANTITY X + 4. SO NOW WE NEED TO MULTIPLY
THIS OUT, AND WE'LL HAVE FOUR PRODUCTS,
ONE, TWO, THREE, AND FOUR. SO WE'LL HAVE X + 6 = X x X,
THAT'S X SQUARED. THEN WE HAVE X x 4, THAT'S 4X. BUT NOTICE HOW THE NEXT PRODUCT
IS ALSO 4X, SO WE'LL HAVE A TOTAL OF 8X,
AND THEN 4 x 4 FOR +16. SO NOW WE HAVE A QUADRATIC
EQUATION THAT WE NEED TO SOLVE. SO WE'LL SET THIS EQUAL TO ZERO
AND SEE IF IT FACTORS. SO TO SET IT EQUAL TO ZERO,
BECAUSE WE HAVE THE SQUARED TERM ON THE RIGHT SIDE, WE'LL SUBTRACT X ON BOTH SIDES, AS WELL AS SUBTRACT SIX
ON BOTH SIDES. SO WE'LL HAVE 0 ON THE LEFT SIDE
= X SQUARED. THIS WILL BE +7X, AND THIS WILL BE +10. SO THIS IS FACTORABLE. THIS WILL FACTOR
TO TWO BINOMIAL FACTORS WITH THE FIRST TERMS WILL BE
THE FACTORS OF X SQUARED, WHICH WILL BE X AND X. AND NOW WE WANT THE FACTORS OF
+10 THAT ADD TO +7, WHICH WILL BE 5 AND 2. SO WE'LL HAVE X + 5 AND X + 2. SO THIS PRODUCT WILL BE 0
WHEN X + 5 = 0 OR X + 2 = 0. SO WE HAVE X = -5 OR X = -2. NOW, TO CHECK THESE, WE'LL SUB THEM BACK
INTO THE ORIGINAL EQUATION TO MAKE SURE THEY ACTUALLY WORK. SO WHEN X IS -5, NOTICE HOW WE'D
HAVE THE SQUARE ROOT OF -5 + 6 THAT WOULD BE THE SQUARE ROOT
OF 1 = ON THE RIGHT SIDE WE HAVE
-5 + 4, WHICH IS EQUAL TO -1. WELL, THE SQUARE ROOT OF 1
IS NOT EQUAL TO -1, AND THEREFORE, X = -5
IS NOT A SOLUTION. THIS IS CALLED
AN EXTRANEOUS SOLUTION, ONE THAT WE FOUND ALGEBRAICALLY,
BUT DOES NOT WORK. NOW WE'LL CHECK X = -2. WELL, IF X IS -2, WE'D HAVE
THE SQUARE ROOT OF -2 + 6 THAT WOULD BE THE SQUARE ROOT
OF 4 = ON THE RIGHT SIDE WE'D HAVE
-2 + 4, WHICH IS EQUAL TO 2, WHICH IS TRUE. AND THEREFORE,
OUR SOLUTION IS X = -2. SO WE ONLY HAVE ONE SOLUTION, EVEN THOUGH ALGEBRAICALLY
IT APPEARED THAT WE HAD TWO. THIS IS THE MAIN REASON
WHY IT'S SO IMPORTANT TO CHECK OUR SOLUTIONS WHEN SOLVING MEDICAL EQUATIONS. LETS TAKE A LOOK AT OUR SECOND
EXAMPLE ON THE NEXT SLIDE. WE WANT TO ISOLATE THE SQUARE
ROOT OF THE QUANTITY 4X + 9. SO THE FIRST STEP HERE
IS GOING TO BE TO UNDO THIS -1 BY ADDING 1
TO BOTH SIDES OF THE EQUATION. SO WE'LL HAVE THE SQUARE ROOT
OF 4X + 9 = X + 1. AND NOW BECAUSE WE HAVE A
SQUARE ROOT WITH AN INDEX OF 2, TO UNDO THIS SQUARE ROOT, WE'LL SQUARE BOTH SIDES
OF THE EQUATION. SO ON THE LEFT SIDE
WE JUST HAVE 4X + 9, AND ON THE RIGHT SIDE
WE HAVE QUANTITY X + 1 TIMES THE QUANTITY X + 1. SO I'LL MULTIPLY THIS OUT,
SO I'LL HAVE 4X + 9 EQUALS-- AGAIN HERE WE'D HAVE FOUR
PRODUCTS, ONE, TWO, THREE, FOUR. SO WE'LL HAVE X x X
THAT'S X SQUARED, X x 1 THAT'S 1X, AND 1 x X IS ALSO 1X. SO WE HAVE +2X AND 1 x 1 = 1. SO, AGAIN,
WE GOT A QUADRATIC EQUATION. SO WE'LL SET THIS EQUAL TO ZERO
AND SOLVE FOR X. SO WE'LL HAVE TO SUBTRACT 4X
ON BOTH SIDES, AND ALSO SUBTRACT 9
ON BOTH SIDES. SO WE'LL HAVE 0 = X SQUARED, THIS WILL BE -2X
AND THIS WILL BE -8. LET'S GO AHEAD AND SOLVE THIS
OVER HERE ON THE LEFT AND SEE IF IT'S GOING TO FACTOR. FIRST TERMS WILL BE X AND X. AND NOW WE WANT THE FACTORS
OF -8 THAT ADD TO -2. IT'LL BE A -4 AND +2. SO WE'LL HAVE X - 4 AND X + 2. SO THIS PRODUCT IS EQUAL TO ZERO
WHEN X - 4 = 0 OR WHEN X + 2 = 0. SO HERE WE HAVE A SOLUTION
OF X = 4 OR X = -2. BUT NOW I NEED TO CHECK THESE
TO MAKE SURE THEY ACTUALLY WORK. SO TO CHECK X = 4,
WE'LL SUBSTITUTE 4 FOR X. SO WE'LL HAVE THE SQUARE ROOT
OF 4 x 4 THAT'S 16 + 9 THAT WOULD BE THE SQUARE ROOT
OF 25 - 1 = 4. THIS WILL BE 5 - 1 = 4,
WHICH IS TRUE, SO X = 4 IS A SOLUTION. AND NOW WE'LL CHECK X = -2. SO IF X IS -2, WE'D HAVE
THE SQUARE ROOT OF 4 x -2 THAT'S -8 + 9, THAT'S A
SQUARE ROOT OF 1 - 1 = -2. WELL, THIS WOULD BE 1 - 1 OR 0. 0 DOESN'T = -2,
SO X = -2 IS NOT A SOLUTION. IT'S CALLED
AN EXTRANEOUS SOLUTION. SO WE DO ONLY HAVE ONE SOLUTION
TO THIS EQUATION, IT IS X = 4. I HOPE YOU FOUND THIS HELPFUL.