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What are the roots of the equation x^2-8x+17=0 in sin

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What are the roots of the equation
x^2-8x+17=0 in sin

What are the roots of the equation x^2-8x+17=0 in sin

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LeonardoVeterano · Tutor por 9 anos

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To find the roots of the quadratic equation $x^2 - 8x + 17 = 0$, we can use the quadratic formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 1$, $b = -8$, and $c = 17$. Plugging these values into the quadratic formula, we get:<br /><br />$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(17)}}{2(1)}$<br /><br />$x = \frac{8 \pm \sqrt{64 - 68}}{2}$<br /><br />$x = \frac{8 \pm \sqrt{-4}}{2}$<br /><br />$x = \frac{8 \pm 2i}{2}$<br /><br />$x = 4 \pm i$<br /><br />Therefore, the roots of the equation $x^2 - 8x + 17 = 0$ are $x = 4 + i$ and $x = 4 - i$.
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