How do you find the value of the discriminant and determine the nature of the roots $-4r^2-4r=6$ ?

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Answer 1 · 2017-07-24T10:28:16

See a solution process below:

First, put this equation in standard form:P

$-4r^2 - 4r - color(red)(6) = 6 - color(red)(6)$

$-4r^2 - 4r - 6 = 0$

For $ax^2 + bx + c = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

The discriminate is the portion of the quadratic equation within the radical: $color(blue)(b)^2 - 4color(red)(a)color(green)(c)$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$color(red)(-4)$ for $color(red)(a)$

$color(blue)(-4)$ for $color(blue)(b)$

$color(green)(-6)$ for $color(green)(c)$

Giving:

$color(blue)(-4)^2 - (4 * color(red)(-4) * color(green)(-6))$

$16 - 96$

$-80$

Because the Discriminate is negative you get a complex solution.

How do you find the value of the discriminant and determine the nature of the roots -4r^2-4r=6-4r^2-4r=6?