TS EAMCET 5 May 2018 Shift 1 Solved Paper

Consider the equation,
x2−3ax+14=0
x2+2ax−16=0
Assume α be the common root. Then,
α2−3aα+14=0
α2+2aα−16=0
From Cramer rule,

α2

48a−28a

=

α

14+16

=

1

2a+3a

α2

20a

=

α

30

=

1

5a

The value of α is,
α=

2a

3

and α=

6

a

Then, the value of a is,

2a

3

=

6

a

a=±3
The value of the given expression is,
a4+a2=(±3)4+(±3)2
=90