SOLUTION: the roots 2x^2-3x+c=0 are imaginary if c equals
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Question 369077: the roots 2x^2-3x+c=0 are imaginary if c equals
Found 2 solutions by jsmallt9, CharlesG2:
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
is called the discriminant. Its value will tell you how many roots of what type there are for a quadratic equation:- means that there are two real roots.
- means that there are one real root.
- means that there are two complex roots. And if b = 0 then the complex roots are imaginary roots (i.e. the real part of the complex number will be zero).
The a, b and c for the discriminant come from the general form for quadratic equations: . Looking at your equation, a = 2, b = -3 and c = c. This makes the discriminant:
which simplifies as follows:
If we want complex roots then we want
Solving this I'll start by subtracting 9 from each side:
Then dividing by -8. (Remember that whenever an inequaity is multiplied or divided by any negative number, like we are doing now, the inequality symbol must be reversed! This is why we have a "greater than" all of a sudden.)
NOTE: Since your b is not zero, there is no way to get imaginary roots.
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
the roots 2x^2-3x+c=0 are imaginary if c equals
use the discriminant part of the quadratic formula
where a = 2, b = -3, c = c
the discriminant is b^2 - 4ac
b^2 - 4ac = (-3)^2 - 4(2)c
b^2 - 4ac = 9 - 8c
if 9 - 8c is less than 0 than the roots are imaginary
9 - 8c < 0
-8c < -9
8c > 9, flipped sign since divided by -1
c > 9/8
if c is greater than 9/8
b does not equal 0, so the results will be complex numbers
a complex number has a real part and an imaginary part
so the answers will not be imaginary numbers
(meaning for example in the case a + bi, a = 0,
the bi is an imaginary number)
instead they will be complex numbers
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