SOLUTION: Evaluate the discriminant of each equation. How many real and imaginary solutions does each have? 37. x2 + 5x + 6 = 0 38. 3x2 - 4x + 3 = 0 39. -2x2

SOLUTION: Evaluate the discriminant of each equation. How many real and imaginary solutions does each have? 37. x2 + 5x + 6 = 0 38. 3x2 - 4x + 3 = 0 39. -2x2 - 5x + 4 = 0 40. 16x2 -

Question 177858: Evaluate the discriminant of each equation. How many real and imaginary
solutions does each have?
37. x2 + 5x + 6 = 0
38. 3x2 - 4x + 3 = 0
39. -2x2 - 5x + 4 = 0
40. 16x2 - 8x + 1 = 0
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

The discriminant is that part of the quadratic equation under the radical, so

then the quadratic equation has two real and unequal roots.

If

then the quadratic equation has two real and equal roots. It is also said that the quadratic has one real root with a multiplicity of 2. This is because the

situation only occurs when the quadratic is a perfect square as in

. Each one of the factors translates to a root of the equation, so there are, in fact, two of them; they just happen to be equal.

If

then the quadratic equation has a conjugate pair of complex roots of the form

where i is the imaginary number defined by

. Note that unless the real part of the complex number (the a in

) is zero, the roots are not purely imaginary.

So, for each of your problems, calculate

and evaluate the character of the roots per the definitions above.

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