SOLUTION: When solving quadratic inequalities, how do I determine the direction of the symbol when writing the solutions out? Ex: x^2+x -6<0 The solution is -3<X<2 Please explain determi

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Question 882989: When solving quadratic inequalities, how do I determine the direction of the symbol when writing the solutions out?
Ex: x^2+x -6<0
The solution is -3 (the opposite direction) Hope this makes sense


Found 2 solutions by josgarithmetic, solver91311:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
You are looking for values of the quadratic function, LESS than zero. Think about this visually, like as a graph. The expression is a parabola opening upward, so the vertex is a minimum. The roots will be at y=0 and the vertex will be below the x-axis.

You can factor the left side expression, so finding the roots is simple to do.
x%5E2%2Bx-6=%28x-2%29%28x%2B3%29%3C0
Which tells you that the roots or zeros are at x=2 and x=-3.
The solution is not -3 !

You know you have a parabola opening upward with a minimum BETWEEN x=-3 and x=2.
LOOK at a graph of just the parabola function:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2%2Bx-6%29

The function or expression will be LESS THAN ZERO for x between -3 and positive 2.

SOLUTION: highlight%28highlight%28-3%3Cx%3C2%29%29.
You can also find the solution using the roots as critical points and testing any point in each interval which the critical points form.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


No, what you presented does not make sense. -3 is NOT a solution to



Step 1: Solve the equation:



Since this is a quadratic for which the discriminant is greater than zero (different signs on the lead and constant coefficients guarantee a greater than zero discriminant), you will get two distinct solutions.

Step 2: Plot the two solutions on the -axis, dividing the axis into three intervals.

Step 3: Select a value INSIDE of each interval -- DO NOT use the interval endpoints, that is, don't use the quadratic equation solution values.

Step 4: Evaluate the original inequality for each of the selected values. Values that make the original inequality true indicate that the entire interval from which the value was selected is, or is in part, the solution set for the inequality.

Step 5: Describe the interval or union of intervals determined in step 4.

Regardless of your outcome, a single number is most assuredly NOT the solution to the given inequality. Further, since the number you gave as an answer is one of the solutions of the quadratic equation and the original inequality is described as absolutely less than (rather than less than OR equal), that number is one of the interval endpoints and cannot possibly be part of the solution set.

John

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