- WE WANT TO SOLVE
THE GIVEN RADICAL EQUATION THAT CONTAINS TWO RADICALS. TO DO THIS, YOU WANT TO
FIRST PUT ONE RADICAL ON EACH SIDE OF THE EQUATION. THEN RAISE BOTH SIDES
OF THE EQUATION TO THE POWER OF THE INDEX. ONCE WE DO THIS, WE'LL SIMPLIFY, AND IF THE EQUATION
STILL CONTAINS A RADICAL, WE'LL ISOLATE THE RADICAL
ON ONE SIDE OF THE EQUATION AND RAISE BOTH SIDES TO
THE POWER OF THE INDEX AGAIN. THEN SIMPLIFY AND SOLVE
THE RESULTING EQUATION AND CHECK OUR SOLUTION
OR SOLUTIONS. SO LOOKING AT OUR EQUATION
AGAIN, NOTICE HOW WE HAVE A SQUARE ROOT
ON EACH SIDE OF THE EQUATION, SO THE NEXT STEP
IS GOING TO BE TO SQUARE BOTH SIDES OF THE EQUATION. SO WE'LL SQUARE THIS, AND THEN WE'RE GOING TO SQUARE
THE ENTIRE RIGHT SIDE, WE CANNOT SQUARE EACH TERM. SO ON THE LEFT SIDE, WHEN WE
SQUARE THE SQUARE ROOT OF 2X + 6 THIS SIMPLIFIES NICELY
TO 2X + 6. BUT ON THE RIGHT, WE ACTUALLY HAVE TO
WRITE THIS OUT AND MULTIPLY. SO WE'RE GOING TO HAVE
THE SQUARE ROOT OF X + 4 + 1 x THE SQUARE ROOT OF X + 4 + 1. AND WHEN MULTIPLYING THIS, WE'RE GOING TO HAVE
FOUR PRODUCTS, ONE, TWO, THREE, AND FOUR. SO WE'LL HAVE 2X + 6
= THE SQUARE ROOT OF X + 4 x THE SQUARE ROOT OF X + 4
WOULD JUST BE X + 4. NEXT WE HAVE THE SQUARE ROOT
OF X + 4 x 1, WHICH IS EQUAL TO
THE SQUARE ROOT OF X + 4. AND NOTICE THIS THIRD PRODUCT
WILL ALSO BE THE SQUARE ROOT OF X + 4. SO WE HAVE +2
x THE SQUARE ROOT X + 4, AND THEN 1 x 1 IS = TO 1. SO WE'VE MADE SOME PROGRESS, BUT NOW WE STILL HAVE
ONE SQUARE ROOT TO DEAL WITH. SO NOW WE'RE GOING TO ISOLATE
THIS SQUARE ROOT, AND THEN SQUARE BOTH SIDES
AGAIN. BUT BEFORE WE DO THIS,
LET'S GO AHEAD AND SIMPLIFY THE RIGHT SIDE. YOU HAVE 2X + 6 EQUALS--
WE HAVE X + 4 + 1 WOULD BE 5, SO 5 + 2
x THE SQUARE ROOT OF X + 4. BE CAREFUL NOT TO ADD THIS 4
AND 2 BECAUSE THIS IS 2
IS ATTACHED BY MULTIPLICATION. AND NOW TO START ISOLATING
THE SQUARE ROOT, LET'S GO AHEAD AND SUBTRACT X
ON BOTH SIDES, AS WELL AS SUBTRACT 5
ON BOTH SIDES. SO WE'LL HAVE X,
THIS WILL BE +1 EQUALS-- THIS IS 0, THIS IS 0, SO WE HAVE 2
x THE SQUARE ROOT OF X + 4. LET'S GO AHEAD AND FINISH THIS
UP HERE. SO WE HAVE THE QUANTITY X + 1 =
2 x THE SQUARE ROOT OF X + 4. NOW WE'LL DIVIDE BOTH SIDES
BY 2, SO WE HAVE X + 1 DIVIDED BY 2
= THE SQUARE ROOT X + 4. NOW, TO UNDO THIS SQUARE ROOT, WE HAVE TO SQUARE BOTH SIDES
AGAIN. SO ON THE RIGHT SIDE WE'RE GOING
TO HAVE THE QUANTITY X + 1 x THE QUANTITY X + 1
ALL OVER, 2 SQUARE WOULD BE 4. THIS IS GOING TO BE = TO X + 4. LET'S GO AHEAD AND MULTIPLY OUT
THE NUMERATOR. WE HAVE X SQUARED + X + X
THAT'S +2X, AND THEN +1. SO WE HAVE X SQUARED + 2X + 1
ALL OVER 4 = X + 4. LET'S FINISH THIS
ON THE NEXT SLIDE. LOOKING AT THE LEFT SIDE HERE,
BECAUSE WE'RE DIVIDING BINOMIAL, WE CAN DIVIDE EACH TERM BY 4. SO WE CAN WRITE THIS AS
1/4 X SQUARED + 2X/4 THAT WOULD BE + 1/2X + 1/4 OR ONE-FOURTH = X + 4. NOW, BECAUSE WE HAVE
A QUADRATIC, WE'RE GOING TO SET THIS EQUAL TO
ZERO AND SEE IF IT CAN FACTOR. SO TO SET IT EQUAL TO ZERO,
WE'LL SUBTRACT X ON BOTH SIDES, AND SUBTRACT 4 ON BOTH SIDES. NOTICE ON THE RIGHT SIDE THIS
WOULD BE 0 AND THIS WOULD BE 0. ON THE LEFT SIDE WE HAVE
1/4X SQUARED AND THEN 1/2X - 1X WOULD BE
- 1/2X OR - ONE HALF X. AND THEN FOR 1/4 - 4,
COMMON DENOMINATOR WOULD BE 4. SO WE'D HAVE 1 - 16
THE NUMERATOR AND 4 THE DENOMINATOR THAT'S GOING TO BE - 15/4. AND NOW TO SEE
IF THIS IS GOING TO FACTOR, WE WANT TO CLEAR
THESE FRACTIONS, SO WE HAVE TO MULTIPLY BOTH
SIDES OF THE EQUATION BY FOUR. SO WE'LL HAVE X SQUARED
- 2X - 15 = 0. LOOKS LIKE THIS WILL FACTOR. THE FIRST FACTORS WILL BE
THE FACTORS OF X SQUARED, WHICH WOULD BE (X,X). AND NOW WE WANT THE FACTORS OF
-15 THAT ADD TO -2. WELL, -5 x 3 = -15,
AND -5 + 3 = -2, SO THAT MEANS ONE OF THE FACTORS
WILL BE X - 5, AND ONE FACTOR WILL BE X + 3. SO THIS PRODUCT IS EQUAL TO ZERO
WHEN X - 5 = 0, OR X = 5, OR WHEN X + 3 = 0,
OR WHEN X = -3. SO ALGEBRAICALLY IT APPEARS
WE HAVE TWO SOLUTIONS, BUT WE DO HAVE TO CHECK THESE
TO MAKE SURE THEY ACTUALLY WORK. SO WHEN X = 5,
LOOKING AT OUR EQUATION, WE'D HAVE THE SQUARE ROOT OF
2 x 5 THAT'S 10 + 6 THAT'S THE SQUARE ROOT OF 16 = THE SQUARE ROOT OF 5 + 4 THE SQUARE ROOT OF 9 + 1. SO WE HAVE 4 = 3 + 1
THAT'S TRUE. SO X = 5 IS A SOLUTION. NOW LET'S CHECK X = -3. WHEN X = -3 ON THE LEFT WE'D HAVE THE SQUARE ROOT
OF -6 + 6 THAT'S GOING TO LEAVE
THE SQUARE ROOT OF 0 EQUALS-- AND ON THE RIGHT SIDE WE'D HAVE
THE SQUARE ROOT -3 + 4. THAT'S THE SQUARE ROOT OF 1 + 1. WELL, 0 DOES NOT EQUAL THE
SQUARE ROOT OF 1 + 1 OR 1 + 1, WHICH WOULD BE 2. SO X = -3 IS WHAT WE CALL
AN EXTRANEOUS SOLUTION AND IS NOT A SOLUTION. WE ONLY HAVE ONE SOLUTION,
IT'S X = 5. OKAY, I HOPE
YOU FOUND THIS HELPFUL.