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- WELCOME BACK FOR A SECOND EXAMPLE
OF SOLVING A QUADRATIC INEQUALITY.
HERE WE WANT TO SOLVE 2X SQUARED + 3X - 5 < 0.
THE SOLUTION TO THIS WILL BE ALL THE X VALUES
THAT SATISFY THIS INEQUALITY.
WE'LL FIRST SOLVE THIS BY HAND AND THEN SOLVE IT GRAPHICALLY.
TO SOLVE THIS BY HAND,
THE FIRST STEP IS TO DETERMINE THE X VALUES
FOR WHICH 2X SQUARED + 3X - 5 = 0.
AND THEN WE'LL USE THOSE VALUES
TO FORM INTERVALS ON THE NUMBER LINE
AND THEN TEST EACH INTERVAL
TO SEE WHICH SATISFY THIS INEQUALITY.
SO AGAIN, THE FIRST STEP IS TO SOLVE
THE EQUATION 2X SQUARED + 3X - 5 = 0.
THIS IS FACTORABLE.
THE FACTORS OF 2X SQUARED ARE 2X AND X.
NOW, WE'LL PLACE THE FACTORS OF -5 IN THE SECOND POSITIONS
SO THAT THE SUM OF THE INTER PRODUCT AND OUTER PRODUCT = 3X.
SO FOR THE FACTORS OF -5,
LET'S PUT A 5 HERE AND A -1 HERE OR -1 HERE.
NOTICE HOW THE INTER PRODUCT IS 5X, THE OUTER PRODUCT IS -2X,
WHICH HAS A SUM OF 3X.
THIS IS FACTORED CORRECTLY
SO THE SOLUTIONS OCCUR WHEN 2X + 5 = 0 OR WHEN X -1 = 0.
WELL, HERE WE WOULD SUBTRACT 5 AND DIVIDE BY 2.
X = -5/2, HERE WE HAVE X = 1.
NOW THESE ARE THE VALUES WHERE 2X SQUARED + 3X - 5 = 0.
SO THESE ARE THE TWO VALUES OF X SQUARED.
2X SQUARED + 3X - 5 = 0,
BUT NOTICE WHEN THIS = 0,
0 IS NOT < 0 SO IT DOES NOT SATISFY THIS INEQUALITY.
SO WHAT THAT MEANS IS WHEN WE FORM OUR NUMBER LINE
AND PLOT THESE TWO VALUES TO FORM INTERVALS
WE'RE GOING TO MAKE OPEN POINTS AT THESE VALUES
TO SHOW THESE ARE NOT PART OF THE SOLUTION.
SO LET'S GO AHEAD AND ASSUME -5/2 IS HERE,
MAKE AN OPEN POINT HERE,
AND LET'S SAY 1 IS HERE
AND MAKE AN OPEN POINT HERE AS WELL.
IF INEQUALITY SYMBOL WAS < OR = 0
THEN THESE POINTS WOULD ACTUALLY BE CLOSED
BECAUSE THESE 2 VALUES WOULD BE PART OF THE SOLUTION.
NOTICE BY PLOTTING THESE 2 POINTS
WE'VE NOW FORMED 3 INTERVALS;
1 ON THE LEFT, 1 ON THE MIDDLE, AND 1 ON THE RIGHT.
SO NOW WE'LL PICK A TEST VALUE IN EACH INTERVAL.
IF THE TEST VALUE SATISFIES THE INEQUALITY
THEN THE INTERVAL IS PART OF THE SOLUTION.
IF IT DOESN'T THEN THE INTERVAL IS NOT PART OF THE SOLUTION.
SO NOW WE'LL PICK A TEST VALUE > -5/2 OR -2.5.
LET'S TEST X = -3.
LET'S TEST X = 0 IN THE MIDDLE.
AND LET'S TEST X = 2 ON THE RIGHT.
AND NOW WE'LL PERFORM SUBSTITUTION
INTO OUR INEQUALITY.
SO WHEN X IS -3 WE WOULD HAVE 2 x -3 SQUARED + 3 x -3 - 5;
WE WANT TO KNOW IF THIS IS < 0.
-3 SQUARED IS 9 x 2 IS 18 + THIS WOULD BE -9 - 5
SO 18 + -9 IS 9, 9 - 5 IS 4, 4 IS < 0.
SO THIS IS TRUE, THEREFORE THE ENTIRE INTERVAL
TO THE LEFT OF -5/2 IS TRUE OR PART OF THE SOLUTION.
SO WE'LL GO AHEAD AND GRAPH THIS INTERVAL
BECAUSE IT'S PART OF OUR SOLUTION.
NOW WE'LL TEST X = 0.
WELL, WHEN X = 0 WE WOULD JUST HAVE 0 + 0 - 5 OR -5 < 0.
THIS IS FALSE, -5 IS NOT < 0
WHICH MEANS THIS INTERVAL HERE IS FALSE AS WELL
OR NOT PART OF THE SOLUTION.
AND NOW WE'LL TEST X = 2.
SO WE'LL HAVE 2 x 2 SQUARED + 3 x 2 - 5 < 0.
2 SQUARED IS 4 x 2 IS 8 SO WE'D HAVE 8 + 6 - 5 < 0.
SO THIS WILL BE 14 - 5, WHICH = 9.
9 IS < THAN 0 SO THE INTERVAL ON THE RIGHT
IS ALSO TRUE OR PART OF OUR SOLUTION.
SO THIS GRAPH IS THE SOLUTION TO OUR QUADRATIC INEQUALITY,
BUT IT'S ALSO LIST THAT USING
INTERVAL NOTATION AND INEQUALITIES.
SO AS THEY MOVE RIGHT WE APPROACH POSITIVE INFINITY
AND AS WE MOVE LEFT WE APPROACH NEGATIVE INFINITY.
SO USING INTERVAL NOTATION IT WOULD BE THE INTERVAL
FROM NEGATIVE INFINITY TO -5/2.
IT'S OPEN ON 5/2 OR THE OPEN INTERVAL FROM 1 TO INFINITY.
USING INEQUALITIES WE COULD SAY X IS > -5/2 OR X IS < 1.
NOW, LET'S GO AHEAD AND SOLVE THIS GRAPHICALLY.
TO SOLVE 2X SQUARED + 3X - 5 < 0
WE GRAPH THE FUNCTION F OF X = 2X SQUARED + 3X - 5.
OR IF WE WANT JUST Y = 2X SQUARED + 3X - 5.
SO THIS INEQUALITY IS GOING TO BE < 0.
WHEN THE FUNCTION IS < 0--
WHEN WE TALK ABOUT FUNCTION VALUES
WE'RE TALKING ABOUT Y VALUES.
SO THE Y VALUES WILL BE < 0
WHEN THE FUNCTION IS ABOVE THE X AXIS.
SO NOTICE HOW THE Y VALUES ARE POSITIVE HERE ON THE RIGHT
HENCE THE FUNCTION IS ABOVE THE X AXIS AS WELL AS ON THE LEFT
WHEN THE FUNCTION IS ABOVE THE X AXIS.
BUT REMEMBER WE'RE LOOKING FOR A SOLUTION IN TERMS OF X.
SO ON THE RIGHT THE FUNCTION IS ABOVE THE X AXIS
WHEN X IS < 1.
SO ON THE X AXIS IT WOULD BE THE OPEN INTERVAL
FROM 1 TO INFINITY AND ON THE LEFT
IT WOULD BE THE OPEN INTERVAL FROM -5/2 TO NEGATIVE INFINITY.
AGAIN, WE DON'T INCLUDE THE X INTERCEPTS HERE
BECAUSE WE WANT TO KNOW WHEN IT'S < 0, NOT < OR = 0.
SO WHILE IT DOES TAKE QUITE A BIT OF WORK
TO SOLVE THESE BY HAND;
SOLVING THEM GRAPHICALLY IS VERY STRAIGHT FORWARD
AS LONG AS THE X INTERCEPTS OF THE FUNCTION
ARE REAL AND RATIONAL.
I HOPE YOU FOUND THIS HELPFUL.