 Hello and welcome to the session. In this session we will discuss a question which says that find the range of values of x for which x square plus 6x minus 11 whole upon x plus 3 is less than minus 1 where x being real. Now before starting the solution of this question we should know about a method and that is the method of intervals. Now here in this method in the first step factorize the quality expression whose coefficient of x square is positive and express left hand side of the inequality in the form x minus alpha the whole into x minus beta the whole where alpha is less than beta. And secondly plot the points pi and beta on the number line dividing the number line into three parts. So after plotting the points alpha beta on the number line starting from the very right all the signs plus minus and plus such that the expression x minus alpha the whole into x minus beta the whole is non-negative in the region on the right of beta. Now when x minus alpha the whole into x minus beta the whole is less than 0 then the required range is alpha less than x less than beta as x minus alpha the whole into x minus beta the whole is negative. So the value of x will lie in between alpha and beta which is a negative region. Now this method will work out as a key idea for solving out this question and now this will start with the solution. Now here it is given x square plus 6x minus 11 whole upon x plus 3 is less than minus 1. This implies x square plus 6x minus 11 whole upon x plus 3 plus 1 is less than 0. This implies now on taking the lcm which is x plus 3 here it will be x square plus 6x minus 11 plus x plus 3 is less than 0. Further this implies x square plus 7x minus 8 whole upon x plus 3 is less than 0. Now multiplying by x plus 3 whole square on both sides this implies x square plus 7x minus 8 the whole into x plus 3 whole square whole upon x plus 3 is less than 0. Further this implies x square plus 7x minus 8 the whole into x plus 3 the whole is less than 0. Now we have to factorize this expression by splitting the middle term. So this implies x square plus 8x minus x minus 8 the whole into x plus 3 the whole is less than 0. Now this implies on factorization this will give x plus 8 the whole into x minus 1 the whole into x plus 3 the whole is less than 0. Now putting each factor equal to 0 we get x is equal to minus 8 1 and minus 3. Now by the method of intervals we will plot these points on the number line. So we have plotted the points minus 8 minus 3 and 1 on the number line. Now starting from the right we will put the signs as plus minus plus minus as x plus 8 the whole into x minus 1 the whole into x plus 3 the whole is less than 0. So the required range by using the formula which is given in the key idea if x minus alpha the whole into x minus beta the whole is less than 0. So the range is alpha less than x less than beta. So here the range will be minus infinity is less than x is less than minus 8 or minus 3 is less than x is less than 1. This means that this expression is negative whenever x is lying in between this range. So this is the solution of the given question and that's all for the session. Hope you all have enjoyed the session.