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Sum/product of roots (1 Viewer)

Thread starter 0hNivlek
Start date Aug 11, 2009

0hNivlek

Member

#1

If alpha and beta are the roots of 3x^2 + 5x - 4 = 0, then find without solving the equation, the values of;

The answer is PlusMinus root 73 over 3. <-- sorry cbb using latex l0l...

I though we just -(a + b) therefore -(-5/3)

LOL. I suck at maths yes.

Last edited: Aug 11, 2009

lychnobity

Active Member

#2

So... what's the question, exactly? (I'm dense)

0hNivlek

Member

#3

OH shit i forgot to post that LOL.

If alpha and beta are the roots of 3x^2 + 5x - 4 = 0, then find without solving the equation, the values of;

lychnobity

Active Member

#4

For convenice, I'm using a and b for the roots

(a-b)² = (a+b)² - 4ab

= (-5/3)² - 4(-4/3) = 73/9

.'. (a-b)² = 73/9

a-b = PlusMinus root 73 over 3

jet

Banned

#5

$3x^2 + 5x - 4 = 0 \\ \sum \alpha = \alpha + \beta = \frac{-b}{a} = \frac{-5}{3} \\ \prod \alpha = \alpha\beta = \frac{c}{a} = \frac{-4}{3} \\ \left (\alpha + \beta \right )^2 = \alpha^2 + 2\alpha \beta + \beta^2 \\ \left (\alpha - \beta \right )^2 = \alpha^2 - 2\alpha\beta + \beta^2 \\ \therefore \left (\alpha - \beta \right )^2 = \left (\alpha + \beta \right )^2 - 4\alpha\beta \\ = \frac{25}{9} + \frac{16}{3} \\ = \frac{73}{9} \\ \therefore \alpha - \beta = \pm \frac{\sqrt{73}}{3}$

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