 Hello and welcome to this session. In this session, we will discuss how to solve quadratic inequality in one variable. I will repeat it. Now our quadratic function is a second degree polynomial given by f of x is equal to ax square plus bx plus c where a, b and c are constants and a is not equal to 0. Now there are four types of quadratic inequalities in one variable. First is ax square plus bx plus c is greater than 0. Second is ax square plus bx plus c is less than 0. Third is ax square plus bx plus c is greater than equal to 0. And fourth is ax square plus bx plus c is less than equal to 0. Now to solve every one of the types of the given inequalities algebraically, we follow three steps. In step one, we write the related equation of the given inequality. That means we replace inequality sign by equality sign. Now for these four types, the related equation will be ax square plus bx plus c is equal to 0. Now in step two, we find the critical values of x by solving the related quadratic equation by using factorization or discriminant rule. Now these critical values are the roots of the given quadratic equation. Thus we find the values of x known as critical values. These values will give us the intervals in which the solution of the quadratic inequality will exist. Now in step three, we plot the critical values on the number line and the two points will divide the number line in three regions. Now this blue shaded portion of the number line is one region. This yellow shaded portion of the number line is the second region and this pink shaded portion of the number line is the third region. So the two points will divide the number line in three regions and these three regions will give us three intervals. Then in step four, we test the values from these intervals which satisfy the given inequality and the inequality is true for that value. The interval which satisfies the given inequality will be the solution set of the quadratic inequality. Now in step five, we write the solution set of the given quadratic inequality and solution set can comprise more than one interval also. Now let us discuss an example. Here we have to solve this quadratic inequality that is we have to solve 3 of square minus 16x plus 5 is less than equal to 0. Now let us start with the solution. In step one, we write its related equation and it will be 3x square minus 16x plus 5 is equal to 0 Now in step two, we will solve this equation. For critical values, now we have 3x square minus 16x plus 5 is equal to 0. Now let us factorize it by splitting the middle term. So this implies 3x square minus 15x minus x plus 5 is equal to 0. This further implies, now from these two terms 3x is common so it will be 3x into x minus 5 is equal to 1 and from these two terms taking minus 1 common it will be minus 1 into x minus 5 is equal to 0. Further this implies x minus 5 is equal to 3x minus 1 the whole is equal to 0. So we get x is equal to 5 and x is equal to 1 upon 3. Now we have obtained the critical values. Now in step three, we will plot the obtained critical values on the other line. Now this is the point 1 upon 3 and this is the point 5. So we have plotted the critical values on the other line. Now here since in the given inequality the sign is less than equal to so here we are represented these points by dark dots which shows that we have also included the points x is equal to 1 upon 3 and x is equal to 5. Now here we can see that the points x is equal to 1 upon 3 and x is equal to 5. Divide the number line in three parts. Now this yellow shaded portion of the number line is the first part and in this first part we have the interval x is less than 1 upon 3 then this blue shaded portion is the second part and in the second part we have the interval 1 upon 3 is less than x is less than 5 and this pink shaded portion of this number line represents the third part and in this third part we have the interval x is greater than 5 now in third part we test the values first of all let us take this first interval that is x is less than 1 upon 3 now let us take every point in this interval so let us take x is equal to 0 that lies in this interval now if x is equal to 0 in the interval inequality so we have 3 into 0 square minus 16 into 0 plus 5 is less than equal to 0 which implies 3 into 0 is 0 minus 16 into 0 is 0 plus 5 is less than equal to 0 and this implies 5 is less than equal to 0 which is false as 5 is greater than 0 so this interval is not the solution set after given inequality now let us see the second interval that is 1 upon 3 is less than x is less than 5 now let us take any point in this interval so let x is equal to 1 now we put x is equal to 1 in the given inequality so we have 3 into 1 square minus 16 into 1 plus 5 is less than equal to 0 and this implies now 1 square is 1 and 3 into 1 is 3 minus 16 into 1 is 16 plus 5 is less than equal to 0 which implies now 5 plus 3 is 8 and 8 minus 16 is minus 8 is less than equal to 0 which is true so this interval is the solution set of the given inequality now let us consider the last interval that is x is greater than 5 now let us take any value lying in this interval so let x is equal to 6 now we put x is equal to 6 in the given inequality so we have 3 into 6 square minus 16 into 6 plus 5 is less than equal to 0 which implies now 6 square is 36 and 36 into 3 is 108 minus 16 into 6 is 96 plus 5 is less than equal to 0 further this implies now 108 plus 5 is 113 minus 96 is 17 is less than equal to 0 which is false as 17 is greater than 0 so this interval is not the solution set of the given inequality thus we have only one solution set that is 1 by 3 is less than x is less than 5 now the values x is equal to 1 upon 3 and x is equal to 5 also satisfy the given inequality thus the solution set is 1 upon 3 is less than equal to x is less than equal to 5 or we can say the solution set is the closed interval 1 upon 3 to 5 also you must note that when we have strict inequalities that is when we have sign off greater than or less than then we do not include the critical values in the solution set so in this session we have discussed how to solve quadratic inequality in one variable algebraically and this completes our session hope you all have enjoyed this session