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4.9 S OLVING Q UADRATIC I NEQUALITIES. EXAMPLE 4 Solve a quadratic inequality using a table Solve x 2 + x 6 using a table. SOLUTION Rewrite the inequality.

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Presentation on theme: "4.9 S OLVING Q UADRATIC I NEQUALITIES. EXAMPLE 4 Solve a quadratic inequality using a table Solve x 2 + x 6 using a table. SOLUTION Rewrite the inequality."— Presentation transcript:

1 4.9 S OLVING Q UADRATIC I NEQUALITIES

2 EXAMPLE 4 Solve a quadratic inequality using a table Solve x 2 + x 6 using a table. SOLUTION Rewrite the inequality as x 2 + x – 6 0. Then make a table of values. Notice that x 2 + x – 6 0 when the values of x are between –3 and 2, inclusive. The solution of the inequality is –3 x 2. ANSWER

3 EXAMPLE 5 Solve a quadratic inequality by graphing Solve 2x 2 + x – 4 0 by graphing. SOLUTION The solution consists of the x -values for which the graph of y = 2x 2 + x – 4 lies on or above the x -axis. Find the graphs x -intercepts by letting y = 0 and using the quadratic formula to solve for x. 0 = 2x 2 + x – 4 x = – – 4(2)(–4) 2(2) x = – x 1.19 or x –1.69

4 EXAMPLE 5 Solve a quadratic inequality by graphing Sketch a parabola that opens up and has 1.19 and –1.69 as x -intercepts. The graph lies on or above the x -axis to the left of (and including) x = –1.69 and to the right of (and including) x = The solution of the inequality is approximately x –1.69 or x ANSWER

5 GUIDED PRACTICE for Examples 4 and 5 Solve the inequality 2x 2 + 2x 3 using a table and using a graph. 5. –1.8 x 0.82 ANSWER

6 S OLVING A QUADRATIC INEQUALITY ALGEBRAICALLY

7 EXAMPLE 7 Solve a quadratic inequality algebraically Solve x 2 – 2x > 15 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with =. x 2 – 2x = 15 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3 or x = 5 Write equation that corresponds to original inequality. Write in standard form. Factor. Zero product property

8 EXAMPLE 7 Solve a quadratic inequality algebraically The numbers –3 and 5 are the critical x -values of the inequality x 2 – 2x > 15. Plot –3 and 5 on a number line, using open dots because the values do not satisfy the inequality. The critical x -values partition the number line into three intervals. Test an x -value in each interval to see if it satisfies the inequality. Test x = – 4: Test x = 1: 1 2 – 2(1)= –1 >15 Test x = 6: The solution is x 5. ANSWER (–4) 2 – 2(–4) = 24 > –2(6) = 24 >15

9 Solve the inequality using any method.


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