Quadratic equations?

Question

Can someone show me how a quadratic formula such as this, 3x^2+bx-3=0, can be solved?

Answer 1

Example

x² - bx - 3 = 0

a = 1, b= - 1, c = - 3

Insert the a, b and c values into the Quadratic formula

Quadratic Formula


x = -b ± √b² - 4ac/2a a ≠0

Answer 2

we know that for any quadratic equation, ax^2+bx+c = 0, the solution will be x = [-b +/- sqrt(b^2-4ac)]/(2a). This is called the quadratic formula.

In your case, the only thing that will cause an unreal solution is when b^2-4ac < 0. This would cause a negative even root, which is imaginary and therefore, unreal.

So, set b^2-4(3)(-3) >= 0 and solve. This will be your solution set.

b^2 >= 36

So b = (inf , -6] U [6 , inf)

Answer 3

the method is by completing the square

3x^2-bx +3 =0

multiply by3 to make 3^2 as a square

(3x)^2-bx +9 =0
or (3x)^2 -bx = -9
or(3x)^2-2(3x)(b/6) = -9
to complete square add (b/6)^2 on both sides

(3x-b/6)^2 = b^2/36 -9
take square root of both sided
(3x-b/6) = +/-sqrt(b^2/36-9)

or 3x = b/6+/-sqrt(b^2/36-9)
x = (b/6+/-sqrt(b^2/36-9))/3

Answer 4

x = -[b +/- sqrt(b^2 + 36)] / 6

I'm afraid you won't get any easier formula :)

For instance, if b = 8 then

x = -[8 +/- sqrt(64 + 36)] / 6
... = -[8 +/- 10] / 6
... = -3 OR -1/3

Answer 5

(-b+-sqrt(b^2+36))/6