 Hello and welcome to the session. In this session we will discuss a question which says that solve x square minus 3x minus 10 whole upon 1 minus x is greater than equal to 2. Now let us start with the solution of the given question. Now when we have solved this inequality for this we will follow some steps to find the solution. Now in step 1 it is standard for this we will take 2 on the left hand side so this inequality will be x square minus 3x minus 10 whole upon 1 minus x minus 2 is greater than equal to 0. Now let us combine left hand side into single section for this we will take a same price. Now in the denominator we have 1 minus 3x minus 10 minus 3 into 1 minus x double and this is greater than combined the left hand side into single section. Now let us see so this implies x square minus 3x minus 10 now minus 2 into 1 is minus 2 minus 2 into minus x is plus 2x 1 minus x is greater than equal to 0. Just further implies now combining the left hand side into the numerator we have x square minus 3x plus 2x would be minus x and minus 10 minus 2 is minus 12 whole upon 1 minus x is greater than equal to 0. Now this is put it from now in step 2 we find real zeros of numerator and denominator first of all let us find zeros of numerator for this we will equate this numerator to 0 and factorize so we have x square minus x minus 12 is equal to 0 now let us factorize it so factorizing we have x square minus 4x plus 3x minus 12 is equal to 0 now where we have factorized by splitting the middle term now this further implies now from these two terms x is common so it is x into x minus 4 the whole and from these two terms 3 is common so it will be plus 3 into x minus 4 the whole is equal to 0 and this implies x minus 4 the whole into x plus 3 the whole is equal to 0 now either x minus 4 is equal to 0 or x plus 3 is equal to 0 so this implies x is equal to 4 or x is equal to minus 3 now where we have found zeros of the numerator similarly we will find zeros of denominator now we have denominator 1 minus x let us equate it to 0 so this implies x is equal to 1 now where we have found zero of the denominator and we have already found zeros of numerator thus the critical values are equal to minus 3 1 and 4 now in step 3 plot all real zeros are critical values on number 9 now this is the point minus 3 this is 1 and this is the point 4 now here you can see that we use post plot or we can say this dot circle for the point minus 3 and the point 4 but we have represented 1 by drawing this whole circle or we can say this open dot because 1 is 0 of denominator so it is restriction and will not be included in solution set now here you can see the values minus 3 1 and 4 divide the number line into four intervals this yellow shaded portion represents the open interval minus infinity to minus 3 then this blue shaded portion represents the open interval minus 3 to 1 this blue shaded portion represents the open interval 1 to 4 and this green shaded portion represents the open interval 4 to infinity so we have 4 intervals now in step 4 choose test number in each interval and check the sign of inequality now let us take the first interval that is the open interval minus infinity to minus 3 or we can write it as x is less than minus 3 now let us take any value y in this interval so that x is equal to minus 4 now let us put x is equal to minus 4 in the given inequality so here we have minus 4 whole square minus of minus 4 minus 12 then 1 minus of minus 4 is greater than equal to 0 which implies minus 4 whole square is 16 minus of minus 4 is plus 4 minus 12 then 1 minus of minus 4 is 1 plus 4 is greater than equal to 0 this further implies now 16 plus 4 is 20 and 20 minus 12 is 8 upon 1 plus 4 is 5 is greater than equal to 0 this implies 1.6 is greater than equal to 0 which is true so this interval is the solution such of the given inequality now let us take the second interval that is the open interval minus 3 to 1 or we can write it as minus 3 is less than x is less than 1 so let us take any value in this interval so let x is equal to 0 that lies in the interval now putting x is equal to 0 in the given inequality we get minus 12 is greater than equal to 0 which is false this interval is not the solution such of the given inequality now let us take the third interval that is the open interval 1 to 1 we can write it as 1 is less than x is less than 4 now let us take any value in this interval now let x is equal to 3 now putting x is equal to 3 in the given inequality we get 3 is greater than equal to 0 which is true this interval is the solution such of the given inequality now let us consider the fourth interval that is the open interval 4 to infinity or we can write it as this greater than 4 let us take any value in this interval so let x is equal to 5 now when we put x is equal to 5 in the given inequality we get minus 2 is greater than equal to 0 which is false this interval is not the solution such of the given inequality now instead we construct summing chart now this inequality is in standard form that is in the term p and q is greater than equal to 0 where p and q are non-zero polynomials and this inequality is called rational inequality now on this number line we see that in the open interval minus infinity to minus 3 we have positive sign for the function p and q then in the open interval minus 3 to 1 we have negative sign for the function p and q then in the open interval 1 to 3 we have positive sign for the function p upon q and lastly for the interval that is the open interval 4 to infinity we have negative sign for the function p upon q so we have open interval minus infinity to minus 3 and open interval 1 to 4 in solution set now since x is equal to minus 3 and 4 are included in the solution set because the given inequality is greater than equal to there are 0's of numerator that is minus 3 and 4 are 0's of numerator thus solution set is 7 open interval minus infinity to minus 3 union 7 open interval 1 to 4 so this is the solution set of the given inequality where minus 3 and 7 are included now where we have plotted the solution set on the number line here we have colored that question of the line that is this number line which forms solution of the given inequality now we can also write the solution set as now this portion represents x is less than equal to minus 3 this portion represents 1 is less than x is less than equal to 4 so the solution set is x is less than equal to minus 3 or 1 is less than x is less than equal to solution of the given question that's all for this session hope you all have enjoyed this session