SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; ju

Algebra ->  Real-numbers -> SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; ju      Log On


   



Question 48581: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
2x^2 + x - 1 = 0
4/3x^2 - 2x + 3/4 = 0
2x^2 + 5x + 5 = 0
3z^2 + z - 1 = 0
m^2 + m + 1 = 0

Found 2 solutions by stanbon, AnlytcPhil:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2 + x - 1 = 0
I'll do one and let you do the rest:
1st: Identify a,b,c for the problem:
In this case a=2,b=1,c=-1
2nd: Evaluate b^2-4ac
In this case you get 1^2-4(2)(-1) =1+8=9
3rd: If the value is > there are two real unequal roots
If the value is 0 there are two real equal roots
If the value is <0 there are two uneqal complex roots
In this case, since the value is 9, there are two real unequal roots.
I'll leave the rest to you.
Cheers,
Stan H.

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
use the discriminant to determine the number of solutions
of the quadratic equation, and whether the solutions are 
real or complex. Note: It is not necessary to find the 
roots; just determine the number and types of solutions. 

----------------------------------

Just learn these rules:

Ax² + Bx + C =0 (or any letter besides x)

has the discriminant  B² - 4AC. Calculate it.

1. If the discriminant is 0, there is one real solution.

2. If the discriminant is negative, there are two conjugate
   complex solutions.

3. If the discriminant is positive, there are two real 
   solutions. 
   A. If the discriminant is a perfect square, the two real 
      solutions are rational.
   B. If the discriminant is not a perfect square, the two 
      real solutions are irrational.

----------------

2x² + x - 1 = 0

Give the x a coefficient of 1:

2x² + 1x - 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 2, B = 1, and C = -1

Calculate discriminant     B² - 4AC
                       (1)² - 4(2)(-1)
                          1 + 8
                            9

9 is positive, so there by rule 3 above there are two real
solutions.  9 is also a perfect square, so by rule 3A the 
two real solutions are rational. 

------------------

4/3x² - 2x + 3/4 = 0 

Compare that to

Ax² + Bx + C = 0

A = 4/3, B = -2, and C = 3/4

Calculate discriminant     B² - 4AC
                       (-2)² - 4(4/3)(3/4)
                           -4
                            

-4 is negative, so there by rule 2 above there are 
two conjugate complex solutions.

-----------------------

2x² + 5x + 5 = 0 

Compare that to

Ax² + Bx + C = 0

A = 2, B = 5, and C = 5

Calculate discriminant     B² - 4AC
                        (5)² - 4(2)(5)
                          25 - 40
                            -13

-13 is negative, so there by rule 2 above there are 
two conjugate complex solutions.

------------------------

3z² + z - 1 = 0

Give the z a coefficient of 1

3z² + 1z - 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 3, B = 1, and C = -1

Calculate discriminant     B² - 4AC
                       (1)² - 4(3)(-1)
                          1 + 12
                            13

13 is positive, so there by rule 3 above there are two real
solutions.  13 is not a perfect square, so by rule 3B the 
two real solutions are irrational. 

-----------------------------------

m² + m + 1 = 0

Give the m² and the m coeficients of 1 each

1m² + 1m + 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 1, B = 1, and C = 1

Calculate discriminant     B² - 4AC
                        (1)² - 4(1)(1)
                           1 - 4
                            -3

-3 is negative, so there by rule 2 above there are 
two conjugate complex solutions.

Edwin