 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says if a and b are the roots of x square minus 3x plus b is equal to 0 and c and d are roots of x square minus 12x plus q is equal to 0 where a, b, c, d form a gp proved that q plus b is to q minus b is equal to 17 is to 15. So first let us learn that if we have a quadratic equation of the form ax square plus bx plus c is equal to 0 tan beta are roots of this equation then sum of roots that is alpha plus beta is given by minus p upon a and product of roots is given by c upon a. So with the help of this idea we are going to solve the above problem so this is our key idea. Let us now begin with the solution. Now here we are given a and b roots of equation x square minus 3x plus p is equal to 0. So by our key idea this implies that a plus b is equal to 3 and a into b is equal to p plus b equation number 1. Now also we are given that c and d are roots of x square minus 12x plus q is equal to 0. So again by our key idea this implies that c plus d is equal to 12 and the product of roots 3 into d is equal to q. So this is the first point we are given this one is second and third we are also given that a, b, c and d are in gp. So this implies b upon a is equal to c upon b is equal to d upon c and let these three ratios be equal to r. So this implies b is equal to a r, c is equal to b r and d is equal to c r. So let this be equation number 2 and these be equation numbers 3. Now this further implies that c is equal to a r square and d is equal to a r cube and we have to show that q plus p is to q minus p is equal to 17 is to 15. So we have q plus p upon q minus p is equal to c d plus a b upon c d minus a b. So we have c is equal to a r square d is a r cube plus a into b is a r upon in the denominator we have a r square into a r cube minus a into a r. This is already equal to a square r raised to the power 5 plus a square r upon a square r raised to the power 5 minus a square r. Now taking a square r common we have r raised to the power 4 plus 1 upon a square r into r raised to the power 4 minus 1. On cancelling we have r raised to the power 4 plus 1 upon r raised to the power 4 minus 1. So this is the value of q plus p upon q minus p. Now we will find the value of r to get the value of this ratio. So let this be equation number 4. From equation 1 and 2 we have a plus b is equal to 3 and c plus d is equal to 12. Now a as it is, b is equal to ar is equal to 3 and c is ar square and d is ar cube. So this is equal to 12. Now let us divide a plus b upon c plus d. So this is equal to 3 upon 12 or we have a plus ar upon. Ar square plus ar cube is equal to 3 upon 12 or taking a common we have 1 plus r and from the denominator taking ar square common we have 1 plus r is equal to 1 upon 4 or this implies 1 upon r square is equal to 1 upon 4 or r is equal to plus minus 2. Now substituting r is equal to 2 in equation number 4 which is q plus p upon q minus p is equal to r raised to the power 4 plus 1 upon r raised to the power 4 minus 1. So we have 2 raised to the power 4 plus 1 upon 2 raised to the power 4 minus 1. So this is equal to 17 upon 15 Now substituting r is equal to minus 2 in equation 4 again we have q plus p upon q minus p is equal to r raised to the power 4 plus 1 upon r raised to the power 4 minus 1. So we have minus 2 raised to the power 4 plus 1 upon minus 2 raised to the power 4 minus 1. So this is again equal to 17 upon 15. Therefore ratio q plus p is equal to 17 is to 15. So this completes the session. Take care and bye for now.