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| Question 76424:  Can you help me with this please??
 When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminant is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. )
 Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant.  Then, graph the corresponding equation.
 
 Answer by Edwin McCravy(20064)
      (Show Source): 
You can put this solution on YOUR website! 
When using the quadratic formula to solve a quadratic 
equation  , the discriminant is  . This discriminant can be positive, 
zero, or negative. (When the discriminant is negative, 
then we have the square root of a negative number. 
This is called an imaginary number, sqrt(-1) = i. )
Explain what the value of the discriminant means to 
the graph of y = ax2 + bx + c. Hint: Chose values of 
a, b and c to create a particular discriminant.  Then, 
graph the corresponding equation?
Choose a = 1, b = 4, c = -21
Then the equation is y = x² + 4x - 21 and the 
discriminant is
(4)² - 4(1)(-21) = 16 + 84 = 100, which is positive.  
The graph intersects the x-axis twice, once at 
x = -7 and again at x = 3. There are two real 
zeros, -7, and 3  -----------------
Now choose a = 1, b = 4, c = 4
Then the equation is y = x² + 4x + 4 and the 
discriminant is (4)² - 4(1)(4) = 16 - 16 = 0.  
The graph just touches the x-axis at -2.  
There is just one real zeros, -2.  [This zero 
is said to have multiplicity 2 because people 
like to think of the graph as "crossing the 
x-axis twice at the same point", and "both its 
two zeros are the same, i.e., 'merging' into
one".]  -------------------
Finally choose a = 1, b = 4, c = 6
Then the equation is y = x² + 4x + 6 and the 
discriminant is (4)² - 4(1)(6) = 16 - 24 = -8, 
which is negative.  The graph does not cross 
or touch the x-axis.  Therefore it has no real
zeros, which means that both its solutions are 
imaginary.  Edwin 
 
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