SOLUTION: solve equation using imaginary and complex numbers x^2+4x+24=0 please help! =)

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Question 140640: solve equation using imaginary and complex numbers x^2+4x+24=0
please help! =)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve x%5E2%2B4%2Ax%2B24=0 ( notice a=1, b=4, and c=24)




x+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A1%2A24+%29%29%2F%282%2A1%29 Plug in a=1, b=4, and c=24



x+=+%28-4+%2B-+sqrt%28+16-4%2A1%2A24+%29%29%2F%282%2A1%29 Square 4 to get 16



x+=+%28-4+%2B-+sqrt%28+16%2B-96+%29%29%2F%282%2A1%29 Multiply -4%2A24%2A1 to get -96



x+=+%28-4+%2B-+sqrt%28+-80+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-4+%2B-+4%2Ai%2Asqrt%285%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-4+%2B-+4%2Ai%2Asqrt%285%29%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=-2+%2B+2%2Asqrt%285%29i or x=-2+-+2%2Asqrt%285%29i