Quadratic Formula Abnormality?

Question

Yes I know the formula -b+-sqrt(b^2-4ac)/2a

but since a 2nd degree equation will go like Ax^2+ Bx +C

now if B^2 is less that -4AC that you end up with a negative under the square root sign. How do you fix this problem.

Should you just normaly go b^2is less than -4ac= square root of absolute value of that answer

i know what you guys are saying where the imaginary number can be such as sqrt(-1) and -sqrt(-1) but that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions.

Answer 1

Consider a = 1, b = 0, c = 1. Clearly, b^2-4ac = -4 < 0. This corresponds to the equation x^2 + 1 = 0. But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.

If you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.

You clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."

Rational equations do indeed give complex solutions, many of them nonreal. The example I gave above (x^2 + 1 = 0) is an example. The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers. If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.

Answer 2

You don't really fix the problem, it means that the solutions are complex numbers. That's why we need complex numbers.

simple ex:
x^2+1 =0
here a=1, b=0, c=1, hence b^-4ac = -4
the QF give the solution as
+sqrt(-4)/2 = +i
and
-sqrt(-4)/2 = -i

so the solutions are quite properly plus and minus the squre root of -1.

Btw, it shows that you are thinking to ask such a question. You will go far as long as you continue to be unafraid to use your mind like that.

Answer 3

That would mean the graph of the quadratic funstion does not touch/cross the x-axis.
You therefore can't solve the equation.
Like the other poster said there are imaginary numbers to address the square root of negative numbers. You'll learn this later.

Answer 4

3x² = 12x - 7 Move everything to the LHS. 3x² - 12x + 7 = 0 Given: ax² + bx + c = 0 Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a Given: 3x² - 12x + 7 = 0 Means: a = 3, b = -12, c = 7 Apply the quadratic formula. x = [-b ± √(b² - 4ac)] / 2a x = [-(-12) ± √((-12)² - 4(3)(7))] / 2(3) x = [12 ± √(144 - 84)] / 6 x = [12 ± √60] / 6 x = [12 ± √(4 * 15)] / 6 x = [12 ± 2√15] / 6 x = [6 ± √15] / 3 ANSWER: x = [6 ± √15] / 3

Answer 5

In that case, you're roots would be imaginary numbers. You can't just take the square root of the absolute value.

Answer 6

answer is an imaginary number sqrt(negative number)

i=sqrt(-1)

all quadratics have solutions, sometimes they are just imaginary

Answer 7

A negative square root does not exist, the function can't be factorised

Answer 8

you fix this "problem" by using imaginary numbers, you'll probably learn about them later