If -1+i is a root of the equation x^4+4x^3+5x | vedclass

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(B) Given,one root is $-1+i$. Since the coefficients of the polynomial are real,the complex conjugate $-1-i$ must also be a root. Thus,$(x - (-1+i))(x

Vedclass · Accepted Answer

$-1 \pm \sqrt{2}$

Vedclass · Answer

(B) Given,one root is $-1+i$. Since the coefficients of the polynomial are real,the complex conjugate $-1-i$ must also be a root. Thus,$(x - (-1+i))(x - (-1-i)) = (x+1-i)(x+1+i) = (x+1)^2 - i^2 = x^2+2x+2$ is a factor of the given equation. Dividing $x^4+4x^3+5x^2+2x-2$ by $x^2+2x+2$,we get the quotient $x^2+2x-1$. Setting $x^2+2x-1 = 0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$. Therefore,the real roots are $-1 \pm \sqrt{2}$.

If $-1+i$ is a root of the equation $x^4+4x^3+5x^2+2x-2=0$,then the real roots of this equation are

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