Solve Lesson 32 Page 57 Math 10 – Kite >

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Topic

Find the intersection of the sets of solutions of two inequalities \( – 3{x^2} + 7x + 10 \ge 0\) and \( – 2{x^2} – 9x + 11 > 0\)

Solution method – See details

Solve two inequalities and combine solutions

Detailed explanation

+ The quadratic triangle \( – 3{x^2} + 7x + 10\) has two solutions \({x_1} = – 1;{x_2} = \frac{{10}}{3}\) and has coefficient \(a = – 3 < 0\)

Using the sign theorem of quadratic triangles, we see that the set of values ​​of \(x\) such that the triangle \( – 3{x^2} + 7x + 10\) bears the sign “+” is \(\left[ { – 1;\frac{{10}}{3}} \right]\)

So the solution set of the inequality \( – 3{x^2} + 7x + 10 \ge 0\) is \(\left[ { – 1;\frac{{10}}{3}} \right]\)

+ The quadratic triangle \( – 2{x^2} – 9x + 11\) has two solutions \({x_1} = – \frac{{11}}{2};{x_2} = 1\) and has coefficient \(a = – 2 < 0\)

Using the sign theorem of quadratic triangles, we see that the set of values ​​of \(x\) such that the triangle \( – 2{x^2} – 9x + 11\) bears the sign “+” is \(\left( { – \frac{{11}}{2};1} \right)\)

So the solution set of the inequality \( – 2{x^2} – 9x + 11 > 0\) is \(\left( { – \frac{{11}}{2};1} \right)\)

Combine two solution sets \(\left[ { – 1;\frac{{10}}{3}} \right]\) and \(\left( { – \frac{{11}}{2};1} \right)\), we have the set of solutions of two inequalities \( – 3{x^2} + 7x + 10 \ge 0\) and \( – 2{x^2} – 9x + 11 > 0\) are \(\left[ { – 1;\frac{{10}}{3}} \right] \cap \left( { – \frac{{11}}{2};1} \right) = \left[{–1;1}\right)\)[{–1;1}\right)\)

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