 Hello and welcome to the session. In this session we will discuss how to apply quadratic equations and equations in one variable to solve problem. Now using quadratic equation let us see the parallel illustration. Now, lava coming from the eruption of volcano follows a parabolic path the height h in feet of a piece of lava. Two seconds after it is ejected from the volcano is given by h of t is equal to minus t square plus 16 t plus 906. Now we have to find after how many seconds does the lava reach height of 1000 feet. Now let us start with its solution. Now where we are given the function of eruption which is a quadratic equation where height h is the function of time t. Let this be equation 1. Now we have to find time t when height h is 1000 feet. So let us put h is equal to 1000 in equation 1. So here we have 1000 is equal to minus t square plus 16 t plus 936. Now let us subtract 936 from both sides. It should be 1000 minus 936 is equal to minus t square plus 16 t plus 936 minus 936. Now forever this will be 1000 minus 936 is 64 is equal to minus t square plus 16 t. Now subtract 64 from both sides of this equation. So it will be 64 minus 64 is equal to minus t square plus 16 t minus 64 this further implies 0 is equal to minus t square plus 16 t minus 64. Now multiplying by minus 1 on both sides we have minus 1 into 0 is equal to minus 1 into minus t square plus 16 t minus 64 the whole. And this implies 0 is equal to minus t square minus 16 t plus 64 or we can write it as t square minus 16 t plus 64 is equal to 0. Now this is a quadratic equation in t. So let us factorize it by splitting the middle term this implies t square minus 8 t minus 8 t plus 64 is equal to 0 and this further implies t into t minus 8 the whole minus 8 into t minus 8 the whole is equal to 0 which further implies t minus 8 the whole into t minus 8 the whole is equal to 0. So either t minus 8 is equal to 0 or t minus 8 is equal to 0 which implies t is equal to 8 or t is equal to 8. So in both these cases we are getting t is equal to 8 thus t is equal to 8 seconds. So if we reach is height of 1000 feet in 8 seconds now many times we use quality inequalities or quality equations in solving the real life problems which are related to find profit, loss, maximum height and minimum height of an object etc. Now we can find the solution of quality inequalities graphically as well as algebraically. Now here we will discuss to find the solution algebraically. Now let us see an example and here calculate daily profit p in dollars for solving x for solving subscriptions is written by the formula p is equal to minus x square plus 8 t x minus 1500 and here we will have to find that for what function of x is hyperbolic positive. Now let us start with the solution. Now here we want to find values of x for which profit that is function of x given by p of x is positive. Now we know that it is positive if it is strictly greater than 0 is positive p of x is greater than 0. Now here profit p of x is determined by the equation minus x 50 x minus 1500. Now p of x is positive if p of x is greater than 0. So here p of x is positive if minus x square plus 8 t x minus 1500 is greater than 0 and this is the quality inequality. Now to find its solution we will follow three steps. In the first step we write its related equation and its related equation will be minus x square plus 8 t x minus 1500 is equal to 0. Now in second step we solve this quadratic equation for critical values of x which will give us the set of intervals which represents the solution of the inequality. Now let us see equation number 2. First of all we need coefficient of x square positive for this we multiply this equation number 2 by minus 1 on both sides and it will be minus 1 into minus x square plus 8 t x minus 1500 is equal to minus 1 into 0 which implies x square minus 8 t x plus 1500 is equal to 0. Now let us factorize this quadratic equation in x by splitting the middle so this implies x square minus 30 x minus 50 x plus 1500 which is equal to 0 which further implies x into x minus 30 the whole minus 50 into x minus 30 the whole is equal to 0 and this implies x minus 30 the whole into x minus 50 the whole is equal to 0. Now either x minus 30 is equal to 0 or x minus 50 is equal to 0 which implies either x is equal to 30 or x is equal to 50 so we have got x is equal to 30 or x is equal to 50. Now in the next step we test for any value from the given intervals and we will check which value satisfies the given inequality that is we will plot these critical points which we have obtained in step 2 on the number 9. Now here we have got x is equal to 30 and x is equal to 50. Now here you can see that the points x is equal to 30 and x is equal to 50 divide the number line in 3 parts. Now in first part that is the split shaded portion we have interval s is less than 30. The second part gives us an interval 30 if less than x is less than 50 that is this yellow shaded portion and the third part gives us interval x is greater than 50 that is this green shaded portion that is like each interval which satisfies the given inequality. First of all we consider the interval s is less than 30 it means all these values on the number line that are less than 30 so let us take every part in this interval let it be x is equal to 29 which is less than 30. Now let us put this value in inequality minus x square plus 80 x minus 1500 is greater than 0 so this will be minus of 29 square plus 80 into 29 minus 1500 is greater than 0. This implies now minus of 29 square now 29 square is 841 plus 80 into 29 is 2320 minus 1500 is greater than 0. Now minus 841 minus 1500 is minus 2341 plus 2320 is greater than 0 and this further implies minus 21 is greater than 0 which is plus s minus 21 is less than 0. Thus this interval is not the solution of the given inequality. Now let us take the second interval that is 30 is less than x is less than 50 it means the values that are between 30 and 50. Now let us take x is equal to 40 that lies in this interval. Now let us put this in given inequality so we have minus of 30 square plus into 30 minus 1500 is greater than 0. And this implies minus 1600 plus 3200 minus 1500 is greater than 0 which implies minus 3100 plus 3200 is greater than 0 and this gives 100 is greater than 0 which is true. This means this interval is the solution set of the given inequality. Now lastly we have this third interval that is x is greater than 50 it means all the values that are greater than 50 so let us take x is equal to 51. Now let us put this value in the given inequality so we have minus of 51 square plus 80 into 51 minus 1500 is greater than 0. This implies minus 2600 and 1 plus 4000 and 80 minus 1500 is greater than 0 this further implies minus 4100 and 1 plus 4000 and 80 is greater than 0. This further implies minus 21 is greater than 0 which is false thus this interval is not the solution set of the given inequality thus solution set is interval 13 is less than x is less than 50 or we can say x belongs to the open interval 30, 50 and solution set is a set of all x such that 13 is less than x is less than 50. Thus Catherine's profit is positive that is p of x is greater than 0 if 13 is less than x is less than 50 that means if x lies between 13 and 50. So in this session we have discussed how to plan a quadratic equations and in equations in one variable to solve real life problems and this completes our session hope you all have enjoyed the session.