- NOW WE'LL LOOK AT SOME MORE EXAMPLES OF SOLVING
QUADRATIC EQUATIONS USING SQUARE ROOTS.
HERE WE HAVE THE QUANTITY X - 5 SQUARED = 16.
BECAUSE THIS QUANTITY INVOLVING X IS A PERFECT SQUARE,
WE CAN SOLVE FOR X BY TAKING THE SQUARE ROOT OF BOTH SIDES
OF THE EQUATION.
SO WE TAKE THE SQUARE ROOT OF THE LEFT SIDE
AND THE SQUARE ROOT OF THE RIGHT SIDE,
AND REMEMBER WE DO HAVE TO INCLUDE A PLUS OR MINUS SIGN
HERE TO MAKE SURE WE OBTAIN OUR TWO SOLUTIONS.
SO THE SQUARE ROOT OF THE QUANTITY X - 5 SQUARED,
SIMPLIFIES TO JUST ONE FACTOR OF X -5.
ON THE RIGHT SIDE WE HAVE +/- THE SQUARE ROOT OF 16,
WHICH SIMPLIFIES TO +/- THE SQUARE ROOT OF 4.
NOW TO SOLVE FOR X, WE'LL ADD 5 TO BOTH SIDES OF THE EQUATION.
WE HAVE X EQUALS--WE NEED TO BE A LITTLE CAREFUL HERE,
THIS WOULD BE 5 +/- 4.
WE DON'T WANT TO LEAVE OUR SOLUTION IN THIS FORM.
WE ACTUALLY WANT TO ADD AND SUBTRACT 4 FROM 5.
SO WE'D HAVE X = 5 + 4, WHICH IS 9 OR X = 5 - 4,
WHICH = 1.
SO WE HAVE TWO REAL RATIONAL SOLUTIONS,
WHICH MEANS IF WE WANTED TO WE COULD HAVE MULTIPLIED
ALL OF THIS OUT, SET IT EQUAL TO ZERO
AND THEN SOLVED BY FACTORING.
OUR NEXT EXAMPLE IS VERY SIMILAR.
WE HAVE THE QUANTITY 3X + 2 SQUARED = 81,
AND SINCE THE LEFT SIDE INVOLVING X IS A PERFECT SQUARE,
WE CAN SOLVE FOR X BY TAKING THE SQUARE ROOT
OF BOTH SIDES OF THE EQUATION.
SO WE HAVE THE SQUARE ROOT OF THE LEFT SIDE,
THE SQUARE ROOT OF THE RIGHT SIDE
AND OF COURSE OUR PLUS OR MINUS SIGN HERE.
SO THE SQUARE ROOT OF THE QUANTITY 3X + 2 SQUARED
SIMPLIFIES TO JUST ONE FACTOR OF 3X + 2.
AND ON THE RIGHT SIDE, SINCE THE SQUARE ROOT OF 81 = 9,
WE HAVE +/- 9.
NOW WE CAN SOLVE THIS EQUATION FOR X.
SO WE'LL START BY SUBTRACTING 2 ON BOTH SIDES.
SO WE'D HAVE 3X EQUALS-- THIS IS GOING TO BE -2 +/- 9
AND THEN DIVIDE BOTH SIDES BY 3.
SO WE HAVE X = -2 +/- 9 ALL OVER 3.
AGAIN, WE DON'T WANT TO LEAVE OUR SOLUTION IN THIS FORM.
WE ACTUALLY WANT TO PERFORM THIS ADDITION AND SUBTRACTION.
SO ONE SOLUTION IS GOING TO BE -2 + 9, WHICH IS 7 DIVIDED BY 3
OR SEVEN-THIRDS.
OR X COULD ALSO BE EQUAL TO -2 -9, THAT'S -11 DIVIDED BY 3.
AGAIN, NOTICE HOW WE HAVE TWO REAL RATIONAL SOLUTIONS,
WHICH MEANS WE COULD HAVE ALSO SOLVED THIS EQUATION
BY FACTORING IF WE MULTIPLIED ALL OF THIS OUT
AND SET IT EQUAL TO ZERO.
AND NOW FOR THE LAST EXAMPLE
WE HAVE THE QUANTITY X + 2 SQUARED = 48.
SO TO SOLVE FOR X WE'LL TAKE
THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION.
WE'D ALSO HAVE A PLUS OR MINUS SIGN HERE.
SO THIS SIMPLIFIES TO ONE FACTOR OF X + 2.
THE RIGHT SIDE IS +/- THE SQUARE ROOT OF 48,
BUT THIS DOES SIMPLIFY BECAUSE 48 DOES CONTAIN
THE PERFECT SQUARE FACTOR OF 16.
SQUARE ROOT OF 48 = THE SQUARE ROOT OF 16 x 3.
THE SQUARE ROOT OF 16 = 4,
SO THIS SIMPLIFIES TO 4 SQUARE ROOT OF 3.
SO WE HAVE X + 2 = +/- 4 SQUARE ROOT OF 3.
AND NOW TO SOLVE FOR X WE'LL SUBTRACT 2
ON BOTH SIDES OF THE EQUATION.
SO WE HAVE X EQUALS--THIS WOULD BE -2 +/- 4 SQUARE ROOT OF 3.
NOTICE HERE WE HAVE TWO REAL IRRATIONAL SOLUTIONS,
WHICH MEANS IF WE MULTIPLIED ALL OF THIS OUT
AND SET IT EQUAL TO ZERO, IT WOULD NOT FACTOR.
AND TO BE CONSISTENT,
LET'S GO AHEAD AND LIST THE TWO SOLUTIONS SEPARATELY.
WE WOULD HAVE X = -2 + 4 SQUARE ROOT OF 3,
AND WE'D HAVE X = -2 - 4 SQUARE ROOT OF 3.
NOW BE CAREFUL HERE,
WE CANNOT ADD AND SUBTRACT 4 FROM THIS 2
BECAUSE THE 4 IS BEING MULTIPLIED BY SQUARE ROOT OF 3.
SO THESE WOULD BE OUR TWO SOLUTIONS.
OKAY, I HOPE YOU FOUND THIS HELPFUL.