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(20054) (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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