Question

The equation whose roots are $2+\sqrt{3},2-\sqrt{3}$, $1+2i,1-2i$, is

• A. $x^4-7x^3-25x^2-43x+40=$
• B. $x^4-7x^3+25x^2+43x-40=0$
• C. $x^4+6x^3+14x^2-22x+5=0$
• D. $x^4-6x^3+14x^2-22x+5=0$

Answer 1

Answer: (D)

$x^4-6x^3+14x^2-22x+5=0$

Roots are $2+\sqrt{3},2-\sqrt{3},1+2i,1-2i$

Quadratic euation of roots $2+\sqrt{3},2-\sqrt{3}$ is

Sum of the roots$=$ $(2+\sqrt{3})+(2-\sqrt{3})=4$

Product of the roots$=$ $(2+\sqrt{3})\ast (2-\sqrt{3})=4-3=1$

thus,quadratic equation is,

$x^2-\left(sumoftheroots\right)x+\left(productoftheroots\right)=0$

i.e, $x^2-4x+1=0$

Quadratic euation of roots $1+2i,1-2i$ is

Sum of the roots$=$ $(1+2i)+(1-2i)=2$

Product of the roots$=$ $(1+2i)(1-2i)=1-{\left(2i\right)}^2=1-4i^2=1+4=5$

thus,quadratic equation is,

$x^2-\left(sumoftheroots\right)x+\left(productoftheroots\right)=0$

i.e $x^2-2x+5=0$

So the equation whose Roots are $2+\sqrt{3},2-\sqrt{3},1+2i,1-2i$ is

$(x^2-4x+1)(x^2-2x+5)=x^4-6x^3+14x^2-22x+5=0$