How do you find the nature of the roots using the discriminant given $9x^2 + 6x + 1 = 0$ ?

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Answer 1 · 2017-11-02T22:34:48

See a solution process below:

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:

$color(red)(9)$ for $color(red)(a)$

$color(blue)(6)$ for $color(blue)(b)$

$color(green)(1)$ for $color(green)(c)$

$color(blue)(6)^2 - (4 * color(red)(9) * color(green)(1)) =>$

$36 - 36 =>$

$0$

Because the Discriminant is $0$ there is just ONE solution for this equation.

How do you find the nature of the roots using the discriminant given 9x^2 + 6x + 1 = 09x^2 + 6x + 1 = 0?