## Solve Lesson 32 Page 57 Math 10 – Kite >

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)\)