Respuesta :
To find the roots of the quadratic equation x^2 + 2x + 5 = 0 is the same as solving it for x.
The formula to get x₁ and x₂ is: x₁,x₂=(-b⁺/₋√(b²-4ac))/(2a) where in our case a=1, b=2 and c=5.
Lets input the numbers:
x₁,x₂= (-2⁺/₋√(2²-4*1*5))/(2*1) = (-2⁺/₋√(4-20))/2, = (-2⁺/₋√(-16))/2
We see that we have a minus sign under the square root so the solutions or roots for our quadratic equation are going to be complex numbers:
x₁ = (-2+4i)/2 = -1+2i
x₂ = (-2-4i)/2 = -1-2i
So our roots are complex and are: x₁= -1+2i and x₂= -1-2i.
The formula to get x₁ and x₂ is: x₁,x₂=(-b⁺/₋√(b²-4ac))/(2a) where in our case a=1, b=2 and c=5.
Lets input the numbers:
x₁,x₂= (-2⁺/₋√(2²-4*1*5))/(2*1) = (-2⁺/₋√(4-20))/2, = (-2⁺/₋√(-16))/2
We see that we have a minus sign under the square root so the solutions or roots for our quadratic equation are going to be complex numbers:
x₁ = (-2+4i)/2 = -1+2i
x₂ = (-2-4i)/2 = -1-2i
So our roots are complex and are: x₁= -1+2i and x₂= -1-2i.
This is the concept of quadratic equations, to get the root of quadratic equation given we proceed as follows;
x^2+2x=-5
Writing the above in the quadratic form ax^2+bx+c=0 we get;
x^2+2x+5=0
(x+1)^2+4=0
x=-1-2i
x=-1+2i
where i=√-1
x^2+2x=-5
Writing the above in the quadratic form ax^2+bx+c=0 we get;
x^2+2x+5=0
(x+1)^2+4=0
x=-1-2i
x=-1+2i
where i=√-1