UPSEE 2017 Solved Paper

Given,
5p2−7p−3=0 .......(i)
5q2−7q−3=0 ........(ii)
It is clear from Eqs. (i) and (ii) p and q satisfy the equation 5x2−7x−3=0
∴p+q=

7

5

and pq=

−3

5

........(iii)
We have to find the equation whose roots are (5p−4q) and (5q−4p)
Here, sum of roots =(5p−4q)+(5q−4p).
=p+q
=

7

5

[from Eq. (iii)]
And product of roots =(5p−4q)(5q−4p)
=81pq−20(p+q)2
=81×(

−3

5

)−20×(

7

5

)2 =

−439

5

∴ Required equation will be x2−( sum of roots) x+ product of roots =0
∴x2−

7

5

x+(−

439

5

)=0
⇒5x2−7x−439=0