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4-2.Quadratic Equations and Inequations
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The roots of the equation ${x^4} - 2{x^3} + x = 380$ are
A
$5, - 4,\frac{{1 \pm 5\sqrt { - 3} }}{2}$
B
$ - 5,4, - \frac{{1 \pm 5\sqrt - 3}}{2}$
C
$5,4,\frac{{ - 1 \pm 5\sqrt - 3}}{2}$
D
$ - 5, - 4,\frac{{1 \pm 5\sqrt - 3}}{2}$
Solution
(a) Given equation is ${x^4} – 2{x^3} + x – 380 = 0$
Using remainder theorem we get
$(x – 5)\,(x + 4)({x^2} – x + 19) = 0$
$x – 5 = 0$, $x – 4 = 0$ and ${x^2} – x + 19 = 0$
$x = 5$, $x = – 4$ and $x = \frac{{ – 1 \pm 5\sqrt { – 3} }}{2}$.
Standard 11
Mathematics
