Complex root of x^2+3x+9=0

Mathematics

1 answer:

Luden [163]3 years ago

6 0

$\text{Quadratic formula}$

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}$

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$x^2+3x+9=0\\\\a=1;\ b=3;\ c=9\\\\\Delta=3^2-4(1)(9)=9-36=-27\\\\\sqrt\Delta=\sqrt{-27}=\sqrt{(27)(-1)}=\sqrt{27}\cdot\sqrt{-1}=\sqrt{9\cdot3}i=\sqrt9\cdot\sqrt3i=3\sqrt3i$

$x_1=\dfrac{-3-3\sqrt3i}{2(1)}=-\dfrac{3}{2}-\dfrac{3\sqrt3}{2}i\\\\x_2=\dfrac{-3+3\sqrt3i}{2(1)}=-\dfrac{3}{2}+\dfrac{3\sqrt3}{2}i$

$i=\sqrt{-1}$

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