AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Given bi-quadratic equation.
f(x)=x4+2x3−16x2−22x+7=0 has one of the root is 2+√3. then one more root will be 2−√3. Now equation whose roots are 2+√3 and 2−√3 is x2−4x+1=0 and so, x2−4x+1 is the factor of bi-quadratic equation
x4+2x3−16x2−22x+7=0
⇒‌‌(x2−4x+1)(x2+6x+7)=0
So, (x2−4x+1) and (x2+6x+7) are factors of given bi-quadratic equation.
Now, the roots of quadratic equation
x2+6x+7=0 aге x=‌

−6±√36−28

2

=−3±√2
∴ The roots of given bi-quadratic equations are
2+√3,2−√3,−3+√2 and −3+√2