What values of b in the following equations will
 give one or more real number solutions?

a). 3x^2+bx-3=0
b). 5x^2+bx+1=0
c). -3x^2+bx-3=0

what is a rule for judging if an equation has 
solutions by looking at it in standard form?

thanks!

When we have an equation of the form 

ax² + bx + c = 0

we form the dicriminant 

Discriminant = b² - 4ac

The rule is

There will be at least one real solution 
if and only if the discriminant is greater 
than or equal 0. There will be one real
solution if the discriminant equals 0 and 
2 real solutions if the discriminant is 
greater than 0.



a).     3x² + bx - 3 = 0

       a = 3, c = -3

       b² - 4ac > 0
  b² - 4(3)(-3) > 0
        b² + 36 > 0
             b² > -36

The square of every real number is
always positive or zero and therefore 
is always greater than -36
             
Answer: b can be any real number.  
   

b). 5x² + bx + 1 = 0

       a = 5, c = 1

       b² - 4ac > 0
   b² - 4(5)(1) > 0
        b² - 20 > 0

(I will use V for the square root radical)
                      __      _
critical values are ±V20 = ±2V5 roughly 4.4 and -4.47

------------o----------------o--------------
          -2V5              2V5

Substituting a test value below -2V5, b=-5
      
     (-5)² - 20 > 0
        25 - 20 > 0
              5 > 0

That's true, so we shade the region
of the number line left of -2V5.

<===========o----------------o--------------
          -2V5              2V5  

Substituting a test value between -2V5 and 2V5, b=0
      
      0² - 20 > 0
       0 - 20 > 0
          -20 > 0

That's false, so we do not shade the region
of the number line between them.

Substituting a test value above 2V5, b=5
      
      (5)² - 20 > 0
        25 - 20 > 0
              5 > 0

That's true, so we shade the region
of the number line right of 2V5.

<===========o----------------o=============>
          -2V5              2V5  

So the values of b for which the solutions are
real are

         (-oo,-2V3) U (2V3,oo)

c). -3x^2+bx-3=0

       a = -3, c = -3

       b² - 4ac > 0
 b² - 4(-3)(-3) > 0
        b² - 36 > 0
 (b - 6)(b + 6) > 0 

                           
critical values are 6 and -6

------------o----------------o--------------
           -6                6 

Substituting a test value below -6, b = -7
      
     (-7)² - 36 > 0
        49 - 36 > 0
             13 > 0

That's true, so we shade the region
of the number line left of -6.

<===========o----------------o--------------
           -6                6  

Substituting a test value between -6 and 6, b=0
      
      0² - 36 > 0
       0 - 36 > 0
          -36 > 0

That's false, so we do not shade the region
of the number line between them.

Substituting a test value above 6, b=7
      
      (7)² - 36 > 0
        49 - 36 > 0
             13 > 0

That's true, so we shade the region
of the number line right of 6.

<===========o----------------o=============>
           -6                6  

So the values of b for which the solutions are
real are

         (-oo,-6) U (2V3,6)
   

Edwin