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Quadratic Inequalities in One variable What are the x-values where the inequality is true

Solving Quadratic Inequalities by Using Algebra (only one variable) Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra. Step 1 Write the related equation. x2 – 10x + 18 = –3

Step 2 Solve the equation for x to find the critical values. Example Continued Step 2 Solve the equation for x to find the critical values. x2 –10x + 21 = 0 Write in standard form. (x – 3)(x – 7) = 0 Factor. x – 3 = 0 or x – 7 = 0 Zero Product Property. x = 3 or x = 7 Solve for x. The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.

Step 3 Test an x-value in each interval. Example 3 Continued Step 3 Test an x-value in each interval. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Critical values Test points x2 – 10x + 18 ≤ –3 (2)2 – 10(2) + 18 ≤ –3 Try x = 2. x (4)2 – 10(4) + 18 ≤ –3  Try x = 4. (8)2 – 10(8) + 18 ≤ –3 x Try x = 8.

Example Continued Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is {x|3 ≤ x ≤ 7} or [3, 7]. –3 –2 –1 0 1 2 3 4 5 6 7 8 9

Another exampl When solving inequalities we are trying to find all possible values of the variable which will make the inequality true. Consider the inequality We are trying to find all the values of x for which the quadratic is greater than zero or positive.

Solving a quadratic inequality We can find the values where the quadratic equals zero by solving the equation, If it is not factorable, we can use the quadratic formula

Solving a quadratic inequality For the quadratic inequality, we found zeros 3 and –2 by solving the equation . Put these values on a number line and we can see three intervals that we will test in the inequality. We will test one value from each interval. -2 3

Solving a quadratic inequality Interval Test Point Evaluate in the inequality True/False

Solving a quadratic inequality You may recall the graph of a quadratic function is a parabola and the values we just found are the zeros or x-intercepts. The graph of is

Solving a quadratic inequality Thus the intervals is the solution set for the quadratic inequality, In summary, one way to solve quadratic inequalities is to find the zeros and test a value from each of the intervals surrounding the zeros to determine which intervals make the inequality true.

Another way is to examine the parabola Standard form ax²+bx+c = y For a graph in ax²+bx+c≥0 form we have to find a solution where the y-value lies on or above the x-axis so we have to find its roots by using quadratic formula…

(For ≤, include the x-intercepts) ax²+bx+c<0 →Graph the related quadratic function which is y=ax²+bx+c and find x-values for which the graph lies below the x-axis. (For ≤, include the x-intercepts) Shade outside Shade inside a<0 a>0 DON’T THINK ABOUT THE NUMBERS ON GRAPH, BUT JUST ITS SHAPE

ax²+bx+c>0 → Graph the related quadratic function which is y=ax²+bx+c and find the x-values for which the graph lies above the x-axis. (For ≥, include the x-intercepts) Shade outside Shade inside a<0 a>0 DON’T THINK ABOUT THE NUMBERS ON GRAPH, BUT JUST ITS SHAPE

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