Quadratic fomula. Right or Wrong?

Question

The problem: 5x^2+3x+1=0

My answer: [-3(+-)(sqr)-11]/10
My question is since negitive 11 is squared is the answer: Not a real number? Because there is no number that will square negitive 11.

Answer 1

Hello dear Neil

5x^2+3x+1=0

● Step 1
if F(x) = ax^2 + bx + c
∆ = b^2 - 4(a)(c) & ∆ ≥ 0
F(x) = 5x^2+3x+1
a = +5 , b = +3 , c = +1
∆ = (3^2) - 4 (5)(1)
∆ = 9 - 20
∆ = -11 < 0
well this function has so real solution

● Step 2
x1 = 1/10 (-3 - i √11)
x2 = 1/10 (-3 + i √11)
so
x 1 = -0.3 - 0.331662i
x 2 = -0.3 + 0.331662i

So you are pretty right .
Good Luck Neil ♣

Answer 2

[-3(+-)(sqrt(-11)]/10 is correct. As you pointed out, the sqrt of a negative number is imaginary. The roots are complex (part real & part imaginary).

Answer 3

If you need a real answer, you have no real roots.

If you are considering complex/imaginary roots, the answers are:
x = -3/10 +/- sqrt(11) i / 10

Remember that i = sqrt(-1).

Answer 4

You are correct! You cannot square out a negative out of the square root.

Answer 5

Yeah you are right, the solutions for the equation are two complex conjugate numbers

Answer 6

Your answer is correct.
Since discriminant is negative
roots will be imaginary
x=[-3+-sqrt(-11)]/10
=-0.3+-i*[sqrt11]/10
where i=sqrt[-1]