What is the discriminant of $9x^2-6x+1=0$ and what does that mean?

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Answer 1 · 2015-07-27T12:01:25

For this quadratic, $Delta = 0$ , which means that the equation has one real root (a repeated root).

The general form of a quadratic equation looks like this

$ax^2 + bx + c = 0$

The discriminant of a quadratic equation is defined as

$Delta = b^2 - 4 * a * c$

In your case, the equation looks like this

$9x^2 - 6x + 1 = 0$ ,

which means that you have

${(a=9), (b=-6), (c=1) :}$

The discriminant will thus be equal to

$Delta = (-6)^2 - 4 * 9 * 1$

$Delta = 36 - 36 = color(green)(0)$

When the discrimiannt is equal to zero, the quadratic will only have one distinct real solution, derived from the general form

$x_(1,2) = (-b +- sqrt(Delta))/(2a) = (-6 +- sqrt(0))/(2a) = color(blue)(-b/(2a))$

In your case, the equation has one distinct real solution equal to

$x_1 = x_2 = -((-6))/(2 * 9) = 6/18 = 1/3$

What is the discriminant of 9x^2-6x+1=09x^2-6x+1=0 and what does that mean?