question!! help?

Question

how do i evaluate the discriminant...
and how do i tell how many salutions has, or whether the salutions are real or imaginary???

1. y = -2x (squere) + 3x - 5


2. y = x(squere)+2x+3



3. y=x(squere)+2x+3

Answer 1

In general, when you have an equation of the form:
y = ax² + bx + c

The quadratic formula is:
x = [ -b ± √ (b² - 4ac) ] / 2a

The descriminant is what is under the sqrt sign in the quadratic formula, namely b² - 4ac.

If the descriminant is 0, then you have 1 real root.
Why? Because when you add or subtract zero you get only one answer, and it is a real number.

If this is positive, you have 2 real roots.
Why? Because the square root of a positive number gives a real answer. The plus/minus gives you two real roots.

If this is negative, then you have 2 imaginary roots.
Why? Because the square root of a negative number gives an imaginary number. The plus/minus gives you two imaginary roots.

In summary:
D = 0 then 1 real root
D > 0 then 2 real roots
D < 0 then 2 imaginary roots

So let's test your equations:
1) y = -2x² + 3x - 5
a = -2, b = 3, c = -5
D = b² - 4ac
D = 3² - 4(-2)(-5)
D = 9 - 40
D = -31
So this would have two imaginary roots

2) y = x² +2x + 3
a = 1, b = 2, c = 3
D = b² - 4ac
D = 1² - 4(2)(3)
D = 1 - 24
D = -23
So this too would have two imaginary roots

3) is the same as #2, so perhaps you copied it wrong...

Answer 2

D= b^2 -4*a*c

1. D= 9 - 4*(-5)*(-2)
D = 9 -40
D= -31 since D<0 the equation has no solution

2. D= 4 - 4*3*1 = 4-12 = -8 doesn't have a solution

3. is the same as 2.


The Discriminant
The quadratic formula can be used when the coefficients are any complex numbers. Now we want to focus our attention to equations with only real number coefficients. Notice the expression under the radical b2 – 4ac in the quadratic formula is called the discriminant. From this number we can determine the nature of the solutions of a quadratic equation.


Three possible cases:



1. b2 – 4ac = 0 Exactly one real number solution exists.


2. b2 – 4ac > 0 Two real number solutions exist.


3. b2 – 4ac < 0 Two complex