- WE ARE GOING TO SOLVE THE GIVEN QUADRATIC EQUATION BY COMPLETING THE SQUARE. COMPLETING THE SQUARE IS ONE OF MANY WAYS TO SOLVE A QUADRATIC EQUATION. THIS VIDEO DOES ASSUME YOU'VE ALREADY WATCHED THE MINI LESSON THAT OUTLINES THE FOLLOWING FIVE STEPS FOR COMPLETING THE SQUARE. IF YOU HAVEN'T DONE SO, YOU MAY WANT TO WATCH THAT VIDEO, OR AT LEAST PAUSE THE VIDEO AND REVIEW THESE STEPS. SO THE FIRST STEP HERE IS TO MOVE THE CONSTANT TO THE RIGHT SIDE OF THE EQUATION. TO DO THIS, WE'LL HAVE TO UNDO THIS 2 BY SUBTRACTING 2 ON BOTH SIDES OF THE EQUATION. THIS WILL GIVE US X SQUARED + 5X. NOW WE ARE GOING TO MAKE THIS A PERFECT SQUARE TRINOMIAL BY ADDING A CONSTANT HERE, SO WE'LL PUT PLUS BLANK EUALS, AND ON THE RIGHT SIDE WE'LL HAVE -2, AND THEN TO MAINTAIN EQUALITY WE'LL ALSO HAVE TO ADD THE SAME CONSTANT THAT WE ADD HERE. SO WE'LL HAVE PLUS BLANK HERE AS WELL. NOW, THE NEXT STEP IS TO MAKE THE LEFT SIDE A PERFECT SQUARE TRINOMIAL. IF YOU LOOK AT OUR NOTES BELOW HERE, IF WE HAVE A QUADRATIC TRINOMIAL WITH A LEADING COEFFICIENT OF 1, THE CONSTANT TERM WOULD NEED TO BE B DIVIDED BY 2 SQUARED, WHERE B IS THE COEFFICIENT OF THE MIDDLE TERM, TO MAKE THIS A PERFECT SQUARE TRINOMIAL. SO IN THIS EXAMPLE NOTICE THAT B IS 5, SO 5 DIVIDED BY 2 SQUARED WOULD BE 5/2 SQUARED, WHICH WOULD BE 25/4. SO WE NEED TO ADD 25/4 HERE ON THE LEFT, AS WELL AS ON THE RIGHT, AND THIS WILL FACTOR INTO A PERFECT SQUARE TRINOMIAL. AND THEN WE NEED TO ADD THE TERMS ON THE RIGHT. TO FACTOR THIS IS A LITTLE BIT MORE CHALLENGING, BECAUSE THE CONSTANT IS A FRACTION. BUT IF WE TAKE A LOOK AT THE NOTES BELOW HERE, THE CONSTANT IN THE BINOMIAL FACTOR IS B DIVIDED BY 2, THE TERM WE HAD HERE BEFORE WE SQUARED. SO IN OUR CASE, B/2 WAS 5/2, WHICH MEANS THIS WILL FACTOR INTO TWO BINOMIAL FACTORS. X AND X ARE THE FACTORS OF X SQUARED, AND THE FACTORS OF 25/4 THAT ADD TO 5 ARE 5/2 AND 5/2. SO WE'LL HAVE PLUS 5/2 HERE AND PLUS 5/2 HERE. AND THEN ON THE RIGHT SIDE -2 = -8/4. OKAY, SO ON THE LEFT WE CAN WRITE THIS AS THE QUANTITY X + 5/2 SQUARED = ON THE RIGHT -8/4 + 25/4 WOULD BE 17/4. AND NOW WE CAN SOLVE FOR X BY UNDOING THE SQUARE BY TAKING THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION. REMEMBER WHEN WE DO THIS WE WOULD HAVE PLUS OR MINUS SIGN HERE TO OBTAIN BOTH SOLUTIONS TO THIS EQUATION. SO ON THE LEFT WE HAVE ONE FACTOR OF X + 5/2, AND ON THE RIGHT WE WOULD HAVE PLUS OR MINUS. THE NUMERATOR IS GOING TO BE THE SQUARE ROOT OF 17. THE DENOMINATOR WOULD BE THE SQUARE ROOT OF 4, WHICH = 2. AND THEN OUR LAST STEP IS TO SUBTRACT 5/2 FROM BOTH SIDES OF THE EQUATION. SO WE HAVE X = -5/2 PLUS OR MINUS SQUARE ROOT OF 17/2. WE ALSO NEED TO RECOGNIZE THAT SQUARE ROOT OF 17 DOES NOT SIMPLIFY, SO IN THIS CASE WE HAVE TWO REAL IRRATIONAL SOLUTIONS. AND BECAUSE WE HAVE A COMMON DENOMINATOR, IF WE WANTED TO WE COULD WRITE THIS AS -5 PLUS OR MINUS SQUARE ROOT OF 17/2. AGAIN WE HAVE TWO SOLUTIONS, BECAUSE ONE SOLUTION IS -5 + SQUARE ROOT OF 17/2, AND ONE SOLUTION IS -5 - SQUARE ROOT OF 17/2. OKAY, WE'LL TAKE A LOOK AT ANOTHER EXAMPLE IN THE NEXT VIDEO.