I'm hoping I did thesse right.
Solve each quadratic equation.
(I don't know how to do symbols on here so i'll write it out in word form)
The equation is: x squared + 3x+4=0 I said its not possible because i got x=3 + or - the square root of -7/2 and you cant find a square root of a negative number.
Am I correct?
Have you ever heard of imaginary roots?
i = sqrt (-1)
sqrt(-7) = sqrt(7) i
no. we havent done them. so your saying you can find the square root of a negative number and solve a quadratic equation?
Yes, you can find the square root of a negative number. In mathematics, we use the concept of imaginary numbers to handle such cases. Imaginary numbers are denoted by the symbol "i," which represents the square root of -1. So, when you have a negative number under the square root, like in the equation x^2 + 3x + 4 = 0, you can express it as √(-7) = √(7)i.
To solve the quadratic equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a, b, and c are the coefficients of the quadratic equation in the standard form, ax^2 + bx + c = 0.
In your case, the equation is x^2 + 3x + 4 = 0, so a = 1, b = 3, and c = 4. Plugging these values into the quadratic formula:
x = (-(3) ± √((3)^2 - 4(1)(4))) / (2(1))
= (-3 ± √(9 - 16)) / 2
= (-3 ± √(-7)) / 2
Since √(-7) can be expressed as √(7)i, the solution becomes:
x = (-3 ± √(7)i) / 2
Therefore, the correct answer to the quadratic equation x^2 + 3x + 4 = 0 is:
x = (-3 + √(7)i) / 2 and x = (-3 - √(7)i) / 2
These are the two complex conjugate solutions to the equation.