JEE Mains 2020 Model Paper 4 New Pattern

Given,
2t2−3t+4 = 0
∴ α + β =

3

2

, α β = 2
Now, sum of roots,
S = (α+

1

α

)+(β+

1

β

)
= (α+β)+(

1

α

+

1

β

)
= (α+β)+(

α+β

αβ

)
=

3

2

+

3

2

×

1

2

=

3

2

+

3

4

=

9

4

Also, product of roots P = (α+

1

α

)(β+

1

β

)
= αβ+(

α

β

+

β

α

)+

1

αβ

⇒ αβ+

α2+β2

αβ

+

1

αβ

= αβ+[

(α+β)2−2αb

αβ

]+

1

αβ

= 2+

( 9 4 −4)

2

+

1

2

=

5

2

+

(9*16)

8

=

5

2

−

7

8

=

13

8

∴ The required quadratic equation is x2−Sx+p = 0
⇒ x2−

9

4

x+

13

8

= 0 ⇒ 8x2−18x+13=0