Question

If $a,b,c,d$ are the roots of the equation $x^4+2x^3+3x^2+4x+5=0$, then $1+a^2+b^2+c^2+d^2$ is equal to

• A. $-2$
• B. $-1$
• C. $2$
• D. $1$`

Answer 1

The correct option is B $-1$

As

$(a+b+c+d{)}^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$

$a^2+b^2+c^2+d^2=(a+b+c+d{)}^2-2(ab+ac+ad+bc+bd+cd)$

$1+a^2+b^2+c^2+d^2=1+(a+b+c+d{)}^2-2(ab+ac+ad+bc+bd+cd)$

Now as $a,b,c,d$ are the roots of the equation, we have

$a+b+c+d=-\frac{2}{1}=-2$ and

$ab+ac+ad+bc+bd+cd=\frac{3}{1}=3$

We have, $1+a^2+b^2+c^2+d^2=1+(-2{)}^2-2(3)=-1$

Hence, option B is correct.