If 1+2i is a root of the equation x^4-3x^3+8x | vedclass

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(A) Given the equation $x^4-3x^3+8x^2-7x+5=0$ with real coefficients,complex roots must occur in conjugate pairs.
Since $1+2i$ is a root,its conjugat

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(A) Given the equation $x^4-3x^3+8x^2-7x+5=0$ with real coefficients,complex roots must occur in conjugate pairs.
Since $1+2i$ is a root,its conjugate $1-2i$ is also a root.
Let the roots be $\alpha, \beta, \gamma, \delta$. Let $\alpha = 1+2i$ and $\beta = 1-2i$.
The sum of the roots $\alpha+\beta+\gamma+\delta = -(-3)/1 = 3$.
$(1+2i) + (1-2i) + \gamma + \delta = 3 \implies 2 + \gamma + \delta = 3 \implies \gamma + \delta = 1$.
The product of the roots $\alpha \beta \gamma \delta = 5/1 = 5$.
$(1+2i)(1-2i) \gamma \delta = 5 \implies (1^2+2^2) \gamma \delta = 5 \implies 5 \gamma \delta = 5 \implies \gamma \delta = 1$.
We need to find the sum of the squares of the other roots,which is $\gamma^2 + \delta^2$.
Using the identity $\gamma^2 + \delta^2 = (\gamma+\delta)^2 - 2\gamma\delta$,we get:
$\gamma^2 + \delta^2 = (1)^2 - 2(1) = 1 - 2 = -1$.

If $1+2i$ is a root of the equation $x^4-3x^3+8x^2-7x+5=0$,then the sum of the squares of the other roots is

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