Verified Answer :

apologiabiology

use quadratic formula

for ax^2+bx+c=0

x=\(\frac{-b+/- \sqrt{b^2-4ac} }{2a}\)

1x^2+2x+9=0
a=1
b=2
c=9
remember
i=√-1 or
i²=-1


x=\(\frac{-2+/- \sqrt{2^2-4(1)(9)} }{2(1)}\)
x=\(\frac{-2+/- \sqrt{4-36} }{2}\)
x=\(\frac{-2+/- \sqrt{-32} }{2}\)
x=\(\frac{-2+/- 4\sqrt{-2} }{2}\)
x=\(\frac{-2+/- 4\sqrt{-2} }{2}\)
x=\(\frac{-2+/- 4i\sqrt{2} }{2}\)
x=\(-1+ 2i\sqrt{2}\) or \(-1- 2i\sqrt{2}\)


SociometricStar

Answer:

\(x=-1+2\sqrt{2}i,\:x=-1-2\sqrt{2}i\)

Step-by-step explanation:

We have been given the quadratic equation \(x^2+2x+9=0\)

The quadratic formula is given by

\(x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Comparing the given equation with \(ax^2+bx+c=0\)

\(a=1,\:b=2,\:c=9\)

Substituting the known values, we get

\(x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \:9}}{2\cdot \:1}\\\\\frac{-2\pm\sqrt{32}i}{2\cdot \:1}\\\\\frac{-2\pm4\sqrt{2}i}{2}\\\\-1\pm2\sqrt{2}i\)

Hence, the values of x are

\(x=-1+2\sqrt{2}i,\:x=-1-2\sqrt{2}i\)

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