 Hello and welcome to the session. In this session we will discuss a question which says that the height of the letter balloon h of t in feet above the ground three seconds after being launched is modelled by the quadratic equation t is equal to minus 16 t square plus 82 t plus 0.25. Now we have to find during what interval the balloon is more than 100 feet above the ground. Now let us start with the solution of the given question. Now here we are given the height of the water balloon h of t in feet above the ground is modelled by the equation that is this quadratic equation and we have to find the interval during which the height of the balloon is more than 100 feet above the ground. It means height should be greater than 100 that is h of t is greater than 100. Now given that h of t is determined by the quadratic equation that is h of t is equal to minus 16 t square plus 82 t plus 0.25. Now here h of t is greater than 100 this means minus 16 t square plus 82 t plus 0.25 is greater than 100. So we have this quadratic inequality. Now we have to find interval for t. So we will solve this inequality for t. Now in the first step we will write its related equation that is minus 16 t square plus 82 t plus 0.25 is equal to 100. Now in the second step we will solve this quadratic equation for critical values of t. Now we want the equation minus 16 t square plus 82 t plus 0.25 is equal to 100. Now we subtract 100 from both sides of this equation. So we have minus 16 t square plus 82 t plus 0.25 minus 100 is equal to 100 minus 100 which implies minus 16 t square plus 82 t minus 99.25 is equal to 0. Now we will multiply both sides of this equation by minus 1. So we have minus 1 into minus 16 t square plus 82 t minus 99.75 the whole is equal to minus 1 into 0 which implies 16 t square minus 82 t plus 99.75 is equal to 0. Now this is a quadratic equation in t. So let us solve this quadratic equation for t. So by using formula we have solved this quadratic equation. Now solving this quadratic equation we get t is equal to 3.14, t is equal to 1.99. Now in step 3 we will plot the critical values on the number line. Now here we have plotted the points 1.99 and 3.14 on the number line. Now here we can see that the points t is equal to 1.99 and t is equal to 3.14 divide the number line in three parts. First part is this yellow shaded portion and in first part we have the interval t is less than 1.99. Now the second part that is this blue shaded portion gives us the interval 1.99 is less than t is less than 3.14. Then the third part that is this green shaded portion gives us the interval t is greater than 3.14. Now we check for region interval which satisfies the given inequality. Now this is the given inequality. Now simply find this inequality by subtracting 100 on both sides we get minus 16 v square plus 82t minus 99.75 is greater than 0. Now let us take the first interval that is t is less than 1.99. Now let us take any point in this interval. So let us take t is equal to 1 that lies in this interval. Now we will put t is equal to 1 in the given inequality. So we will have minus 16 into 1 square plus 82 into 1 minus 99.75 is greater than 0. This implies minus 16 plus 82 minus 99.75 is greater than 0. This implies minus 115.75 plus 82 is greater than 0 which implies minus 33.75 is greater than 0 which is false as minus 33.75 is less than 0. So this interval is not the solution set of the given inequality. Now let us take the second interval that is 1.99 is less than t is less than 3.14. Now let us take any value in this interval. So let t is equal to 2 that lies in this interval. Suppose t is equal to 2 in the given inequality. So here we have minus 16 into 2 square plus 82 into 2 minus 99.75 is greater than 0. This implies minus 16 into 2 square now 2 square is 4 and 4 into minus 16 is minus 64 plus 82 into 2 is 164 minus 99.75 is greater than 0. Now this implies minus 163.75 plus 164 is greater than 0 which implies 0.25 is greater than 0 which is true thus this interval is the solution set of the given inequality. Now let us take the third interval that is t is greater than 3.14. Now let us take any value in this interval let t is equal to 4 that lies in this interval. Now put t is equal to 4 in the given inequality. So here we have minus 16 into 4 square plus 82 into 4 minus 99.75 is greater than 0. Now this implies now minus 16 into 2 square now 4 square is 16 and 16 into minus 16 is minus 256 plus 82 into 4 is 328 minus 99.75 is greater than 0. Further this implies minus 355.75 plus 328 is greater than 0 which implies minus 27.75 is greater than 0 which is false as minus 27.75 is less than 0 thus this interval is not the solution set of the given inequality thus the solution set of the given inequality is the interval 1.99 is less than t is less than 3.14 that is the open interval 1.99 to 3.14 Now in the given inequality we have greater than sign it means here the values 1.99 and 3.14 will not be included in the solution set that is why the solution set is the open interval 1.99 to 3.14. Now on the number line we can represent the solution set in this way that is this shaded portion of the number line represents the solution set and we have put hollow circles to represent the points 1.99 and 3.14 which means these two points are not included in the solution set. So the solution set is given by the interval 1.99 is less than t is less than 3.14 thus between 1.99 seconds to 3.14 seconds the height of the loon is greater than 100 feet. So this is the solution of the given question that is all for this session hope you all have enjoyed this session.