UPSC NDA Math Model Paper 2

Given:
(x2+2)2+8x2=6x(x2+2)
Let x2+2=t⇒t2+8x2=6xt
⇒t2−6xt+8x2=0
⇒t2−4xt−2xt+8x2=0
⇒t(t−4x)−2x(t−4x)=0
⇒(t−2x)(t−4x)=0
Now, putting the value of t in above equation, we get
⇒(x2+2−2x)(x2+2−4x)=0
Now, (x2−2x+2)=0 and (x2−4x+2)=0
⇒x2−2x+2=0
x=

−(−2)±√(−2)2−4×1×2

2×1

=

2±√−4

2

=

2±2i

2

=1±i
Roots are imaginary.
Now,
⇒(x2−4x+2)=0
x=−(−4)±

√(−4)2−4×1×2

2×1

=

4±√8

2

=

4±2√2

2

=2±√2
Roots are real.
We can say that not all roots of the equation are complex.So statement 1‌st is wrong.
Now,
Sum of all the roots =(1+i)+(1–i) +(2+√2)+(2−√2)=6
So statement 2‌nd is true.