The roots of the equation x^4 - 2x^3 + x = 38 | vedclass

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(A) Given equation is $x^4 - 2x^3 + x - 380 = 0$. By testing integer roots,we find that $x = 5$ and $x = -4$ are roots. Dividing the polynomial $x^4 -

Vedclass · Accepted Answer

$5, -4, \frac{1 \pm 5\sqrt{-3}}{2}$

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(A) Given equation is $x^4 - 2x^3 + x - 380 = 0$. By testing integer roots,we find that $x = 5$ and $x = -4$ are roots. Dividing the polynomial $x^4 - 2x^3 + x - 380$ by $(x - 5)(x + 4) = x^2 - x - 20$,we get the quotient $x^2 - x + 19$. So,$(x - 5)(x + 4)(x^2 - x + 19) = 0$. Setting $x^2 - x + 19 = 0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $x = \frac{1 \pm \sqrt{1 - 4(19)}}{2} = \frac{1 \pm \sqrt{-75}}{2} = \frac{1 \pm 5\sqrt{-3}}{2}$. Thus,the roots are $5, -4, \frac{1 \pm 5\sqrt{-3}}{2}$.

The roots of the equation $x^4 - 2x^3 + x = 380$ are

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