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Sandstorm
11-22-2010, 01:32 AM
What is a real root?

I have the quadratic equation 3x^2 + 9x - 2 = 0.

I solve it using the quadratic formula and get

X = (-9 +- sqrt(105))/6

The discriminant is 105, which means this equation has two real roots - as far as I know, that is irreducible, and I'm unsure whether that's two real roots or not.

Edit:

WolframAlpha tells me this

x = 1/6 (-9-sqrt(105))

and

x = 1/6 (sqrt(105)-9)

Are these the same as what I have up there?

blulightning
11-22-2010, 01:37 AM
I don't really understand your notation, but here's what I can tell you about Real and Imaginary roots

Real roots are within the real number set, i.e, negative infinity to positive infinity.

Imaginary roots include the number i. For example, the equation x^2= -1. The two roots of the equation are i and -i, or, sqrt(-1) and -sqrt(-1)

Ogre
11-22-2010, 01:40 AM
To be real means to not be complex, meaning it doesn't contain the square root of a negative number or i, and yours doesn't because the discriminant is positive.

What WolframAlpha says is the same as what you have, it's just making the division into multiplication by a fraction.

TomTuff
11-22-2010, 02:08 AM
yes, it is. It just simplifies it into fractions

Sandstorm
11-22-2010, 02:10 AM
I understand multiplying by 1/6 is the same as dividing by 6, what I was wondering is the part after that - it didn't make sense to me. Thanks for replying, I get it now.

TomTuff
11-22-2010, 06:03 AM
I understand multiplying by 1/6 is the same as dividing by 6, what I was wondering is the part after that - it didn't make sense to me. Thanks for replying, I get it now.

it's giving you both roots

Markus
11-22-2010, 03:48 PM
Real root: sqrt(x), x >= 0.
Remember that x² = y, y can be sqrt(x) or -sqrt(x)

Richard
11-22-2010, 11:31 PM
x = 1/6 (-9-sqrt(105))

and

x = 1/6 (sqrt(105)-9)

are the same thing :P It just moved the -9 (mathmaticians prefer to not start statements/equations with a negative number if possible, just looks nicer).

As Markus said, sqrts can be both positive and negative, as negative number squared gives a positive. That said, any even integer root has 2 solutions (square root, quartic root, hexic root etc.), but odd integer roots always keep the same positivity.

If it helps to imagine it on a graph with a quadratic equation, the line goes through the x axis twice (unless the discriminant is 0, in which case it merely touches it, or the discriminant is negative, in which case never, as the entire line is above the x axis), thus has 2 results for x. And then cubic has 3, quartic 4, pentic 5, you get the picture :p

blulightning
11-22-2010, 11:38 PM
I understand multiplying by 1/6 is the same as dividing by 6, what I was wondering is the part after that - it didn't make sense to me. Thanks for replying, I get it now.

You've gotten a zillion responses after posting this, by now you should be an expert at Complex Roots :)

Sandstorm
11-23-2010, 03:47 AM
You've gotten a zillion responses after posting this, by now you should be an expert at Complex Roots :)

Ikr ^_^.

Anyway, thanks for all your replies guys!

Laughinn
03-22-2011, 04:41 AM
Use the trinomial therom , after that u should have (x-?)(x+?) , to find the roots do the opposite of the two numbers you get , very simple if u got 6 and 7 roots are -6 -7