One, Two, or No Real Solution Problem.

Question

How do I know if this has One, Two, or No Real Solution?

 

-3x^2 + 5x - 1 = 0

 

 

Answer 1

3x² -5x+1=0

Quadratic eq ax² +bx+c=0

Discriminant=b² -4ac  

                  =(-5)² -4*3*1

                  =25-12

                  =13>0  so  2 solutions. 

 

Answer 2

You need to find the discriminant which is the amount under the radical of the quadratic formula. Whether this number is positive, negative, or zero determines how many solutions exist.

 

Standard form of Quadratic Equation: ax² + bx + c = 0

 

Quadratic Formula: x = ( -b ± √(b² - 4ac) ) / (2a)

Discriminant: Δ = b² - 4ac

 

If Δ>0  then 2  real solutions

If Δ=0 then 1 double real solution

If Δ<0 then 0 real solutions (but 2 complex solutions)

 

-3x² + 5x - 1 = 0

a=-3

b=5

c=-1

Δ = (5)² - 4(-3)(-1) = 25 - 12 = 13 >0

So, 2 real solutions

which are x = ( -5 ± √13 ) / ((2)(-3) ) = (-5±√13)/-6

x∈ { (5+√13)/6, (5-√13)/6 }

Answer 3

-3x² + 5x - 1 = 0

3x² - 5x + 1 = 0

 

What is the value of the discriminant?