 Hi friends, so welcome again to another problem solving session. So you know problem solving is an integral part of understanding the concept. So the idea is to go through these sessions and then work on the problems given in the worksheet. Now, if alpha and beta are zeros of the quadratic polynomial, this given polynomial is 3x square minus 4x plus 1, and you have to find out another quadratic polynomial whose zeros are this and beta square plus alpha. So it's always easy to simplify this like this, that and you know let's say alpha and beta are, alpha and beta are zeros, zeros of fx, okay? Then we have to find out gx, find gx, gx whose zeros are, whose zeros are, let us say capital A, capital A which is given by alpha square by beta, capital A is given by alpha square bond beta and capital B, capital B, capital B is equal to beta square by alpha, correct? Now, what do we know? First of all, let us understand what does it mean that alpha and beta are zeros of fx. Now fx I'm rewriting is nothing but, what is fx? 3x square minus 4x plus 1, isn't it? 3x square minus 4x plus 1 and if alpha and beta are the zeros, then we know alpha plus beta is nothing but negative of coefficient of x divided by coefficient of, coefficient of x squared, isn't it? Which is equal to negative of negative 4 and divided by 3, again, do not miss the negative sign while taking the coefficient, so hence it is 4 upon 3, correct? 4 upon 3 and what is alpha plus, alpha times beta that is product of the roots is nothing but coefficient of, coefficient of constant term or the constant term itself, let us write constant term. So constant term, constant term divided by coefficient of x squared, so what is constant term here? 1 and coefficient of x square is 3, so hence it is 1 by 3, right? Now if roots of a equation is given, so sorry, or the zeros of a polynomial is given, then polynomial can be expressed as in terms of zeros, so basically we are now trying to express a polynomial, polynomial that to a quadratic polynomial, so I am writing quadratic polynomial in terms of, in terms of its zeros, its zeros is given by nothing but let us say in this case we have taken gx, so gx will be nothing but a constant k times x square minus sum of the roots, sum of the roots times x plus product, product of roots or I should be using zeros instead of roots, so let me write product of sum of the zeros because we are dealing with polynomials and here also product of zeros. And what are the desired products, sorry, desired roots, so gx had roots a and b, capital a and b isn't it, if you remember. What is capital a? Alpha square by beta and capital b was beta square by alpha, so my dear friends, sum of the proposed root or zeros, sum of zeros will be, sum of zeros of the new equation, not the previous equation, sum of zeros of the desired equation or desired polynomial, desired polynomial, sum of the zeros of the desired polynomial is a plus b which is equal to alpha square by beta plus beta square by alpha which is nothing but alpha cube plus beta cube by alpha beta, if you take the LCM you will see you will get this expression. Now do we know alpha beta, indeed we know alpha beta, we definitely know alpha beta why because alpha beta is 1 upon 3, isn't it? But certainly alpha cube plus beta cube we don't know in this form, so let us use quadratic, sorry, identities to convert alpha cube plus beta cube into, we know alpha cube plus beta cube will be alpha plus beta times, or rather let us first explain how to find this out so that it becomes convenient for you. So I can write alpha plus beta whole cube, the identity is alpha cube plus 3 alpha square beta plus 3 alpha beta square plus beta cube, now if you notice alpha cube and beta cube is what we desire, so hence we can now say this as alpha plus beta whole cube is equal to alpha cube plus beta cube plus 3 alpha beta, so if you take alpha beta common and 3 common from these two terms what is left? Alpha plus beta, so hence alpha cube plus beta cube will be equal to alpha plus beta whole cube minus 3 alpha beta alpha plus beta, why minus? Because this particular term goes to this side, so it becomes minus. Now thankfully we know alpha plus beta as well and alpha plus beta alpha beta as well, so let us deploy the values, what do we know alpha plus beta was 4 by 3 if you see here and alpha beta was 1 by 3, so if you notice we have expressed one of the part of, basically we are now trying to find out the sum of the new polynomial, sum of the zeros of the new polynomial and for that we require two values, alpha cube plus beta cube where alpha and beta are the zeros of the given polynomial given in the question and alpha beta is the product of that route. So basically we were trying to find out the value of this and alpha beta is already known, so if we know these two values we will know the sum of the roots of, or sum of the zeros of the new polynomial, so now we know, so alpha plus beta is how much, 4 by 3, so we can write this as 4 by 3 cube minus 3 1 by 3 times 4 by 3, is that it alpha plus beta was 4 by 3, so hence what is this value, 4 cube is 64 and this is 27 minus 4 upon 3 which can be written as, 27 will be the LCM so hence it is 64 and here 9 4s are 36, so this is the value 36 by 27, so then it is 28 upon 27, is it it, 28 upon 27 is alpha cube plus beta cube, now what is the value of a plus b, the sum of zeros of the new polynomial is alpha cube plus beta cube by alpha plus beta, is it it, that's what we needed to find out see here, correct, so let us, or rather let us write here itself, so alpha cube plus beta cube is how much, 28 by 27 and alpha beta was 1 upon 3, 1 upon 3 is it it, where did we find that out, 1 upon 3, here alpha beta was 1 upon 3, so hence what is this now, so 28 by 28, sorry 28 by 9 is the sum of the new zeros and what is a b, product of new zeros is nothing but alpha square by beta into alpha by, or rather than beta square by alpha, can you see that, the two new proposed zeros where a was alpha square b by beta and b was beta square by alpha, so a b we have to find out the product of the new zeros, which is nothing but alpha square by beta and beta square by alpha, which is nothing but alpha beta itself, because this alpha this 2 will go, this beta this 2 will go and when alpha beta is nothing but 1 upon 3, we have just found out above see, alpha beta is 1 by 3, so now we get what, we got alpha plus beta, sorry a plus b which is nothing but 28 by 27 and we got a b which is 1 by 3, so the new equation or the new polynomial will be, sorry not equation, the polynomial is gx was given as k constant, any constant times x square minus sum of the new zeros times x plus product of the new zeros, okay, so gx is equal to k times x square minus a plus b x plus a b, so let's deploy the values, so what is it, k times x square minus alpha plus beta is nothing but, I'm sorry this is alpha plus beta is not 28 by 27, actually it is 28 by 9, so let me just correct it, so this is 28 by 9, because here we found out it is 28 by 9, so 28 by 9, so what is it, so hence we get x square minus 28 by 9x plus 1 upon 3, that's how, so this is the new polynomial k x square minus 28 by 9x plus 1 upon 3 where k is any constant k is any real constant.