How do you find the roots of #x^ { 2} + 3x + 8= 0#?

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Answer 1 · 2018-07-25T01:25:49

The roots are #x = (-3 +- isqrt23)/2#.

#x^2 + 3x + 8 = 0#

To find the roots (solutions of #x#), use the quadratic formula #x = (-b +- sqrt(b^2 - 4ac))/(2a)#.

We know that #a = 1#, #b = 3#, and #c = 8#, so let's plug them into the formula:

#x = (-3 +- sqrt(3^2 - 4(1)(8)))/(2(1))#

#x = (-3 +- sqrt(9 - 32))/2#

#x = (-3 +- sqrt(-23))/2#

Since a square root of a negative number has no real solution, the roots are imaginary.

We know that #i# (imaginary number) is equal to #sqrt(-1)#, so we can take that out and have an imaginary solution:
#x = (-3 +- isqrt23)/2#

Hope this helps!