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If $a, b, c$, and $d$ are the roots of the equation $x^{4}+2 x^{3}+3 x^{2}+4 x+5=0$, then $1+a^{2}+b^{2}+c^{2}+d^{2}$ is equal to
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Verified Answer
The correct answer is:
$-1$
As,
$$
\begin{gathered}
(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2} \\
+2(a b+b c+a d+b c+b d+c d) \\
\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2} \\
-2[a b+b c+a c+a d+b d+c d]
\end{gathered}
$$
Now, as $a, b, c, d$ are the roots of the equation.
We have,
$$
a+b+c+d=\frac{-2}{1}=-2
$$
and $a b+a c+a d+b c+b d+c d=\frac{3}{1}=3$
We have, $1+a^{2}+b^{2}+c^{2}+d^{2}$
$$
\begin{aligned}
&=1+(-2)^{2}-2(3) \\
&=1+4-6=-1
\end{aligned}
$$
$$
\begin{gathered}
(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2} \\
+2(a b+b c+a d+b c+b d+c d) \\
\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2} \\
-2[a b+b c+a c+a d+b d+c d]
\end{gathered}
$$
Now, as $a, b, c, d$ are the roots of the equation.
We have,
$$
a+b+c+d=\frac{-2}{1}=-2
$$
and $a b+a c+a d+b c+b d+c d=\frac{3}{1}=3$
We have, $1+a^{2}+b^{2}+c^{2}+d^{2}$
$$
\begin{aligned}
&=1+(-2)^{2}-2(3) \\
&=1+4-6=-1
\end{aligned}
$$
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