 Hello and welcome to the session. In this session, we will discuss a question which says that if both the rules of the quadratic equation, x square minus p plus 3, the whole, into x plus p plus x is equal to 0, are negative, calculate the values of p. Now before starting the solution of this question, we should know about the method. And that is the method of intervals. For this method, in the first step, we have to factorize the quality expression whose coefficient of x square is positive x plus the left hand side of the inequality in the form minus alpha, the whole, into x minus beta, the whole, where alpha is less than beta. In the second step, plot the points theta and beta on the number line, thus dividing the number line into 3 parts. Now after dividing the number line into 3 parts, and after plotting the points alpha and beta here, starting from the right-most region, put the signs plus minus the expression x minus alpha, the whole, into x minus beta, the whole, is more negative than the right of beta. When x minus alpha, the whole, into x minus beta, the whole, is greater than or equal to 0, then the required range is less than x is less than or equal to alpha, beta is less than or equal to x is less than infinity. Now this method, we work out as a key area, first solving out this question, and now we will start with the solution. Now given the quadratic equation, x square minus p plus 3, the whole, into x plus p plus 6, is equal to 0. Now let, which be minus alpha and minus beta, where, now we have taken these, as the range of this equation. Now comparing this equation, with the same form of quadratic equation, here a is equal to 1, b is equal to minus of p plus 3, and c is equal to p plus 6. Now is equal to minus b over a, that means minus alpha plus of minus beta is equal to minus, but in the values of b and a here, it will be minus of p plus 3, now 1, minus alpha minus beta is equal to p plus 3. In this common written brackets, alpha plus beta is equal to p plus 3. This implies, p plus 3 is less than 0, this is a negative quantity. Also, p over a, that means minus alpha into minus beta is equal to, put in the values of c and a here, it will be, alpha beta is equal to p plus 6. This implies, into beta, take this as 1, it is less than 0, and is greater than 0. This implies, p is greater than minus 6. Now, on the mileage tool, this is less than b, is less than minus 3. Now let us name it as, d is greater than equal to, b square minus 4 a c is greater than equal to 0. In the values of a, b and c here, this implies, minus of p plus 3 minus 4 into 1 into p plus 6, the whole is greater than equal to 0. 6 the whole is greater than equal to 0. On p square, this will be, p square plus 9 plus 6 p minus 4 p minus 4 into 6 is 24 is greater than equal to 0. However, this implies, minus 15 is greater than equal to 0. Now this implies, by splitting the middle term, v3p minus 15 is greater than equal to 0. Minus 3, given like it's p plus 5, is greater than equal to 0. 5 the whole is greater than equal to 0. Now putting, p is equal to 3 and p is equal to minus 5. Now by using the method of intervals, we will plot the points 3 and minus 5 on the number line. 5 and 3 on the number line, now starting from the very right, minus and plus. The whole into p minus 3 the whole is greater than equal to 0. Then according to the result given in the key idea, 6 minus alpha the whole into x minus beta the whole is greater than equal to 0. Then the p is less than x is less than equal to alpha, beta is less than equal to x less than infinity. So the required range is minus infinity is less than p is less than equal to minus 5 is less than equal to p less than infinity. In results minus 6 is less than p is less than equal to minus 5. In equation 3, we have minus 6 less than p less than minus 3. In equation 4, we have minus infinity is less than p is less than equal to minus 5. So combining these two, we have this result. This is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.