Roots of quadratic equation

**maybeline9216** · October 12th 2008, 10:51 PM

Please teach me how to do part b(iii).

**Jhevon** · October 12th 2008, 11:05 PM

Originally Posted by maybeline9216

Please teach me how to do part b(iii).

note that $(\alpha - \beta)^2 = \alpha^2 - 2 \alpha \beta + \beta^2$ . with this, you can find $\alpha - \beta$ by square rooting both sides.

but how to find $\alpha^2 - 2 \alpha \beta + \beta^2$ ?

note that $(\alpha + \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2$

$\Rightarrow (\alpha + \beta)^2 - 4 \alpha \beta = \color{red}\alpha^2 - 2 \alpha \beta + \beta^2$

and you know $\alpha + \beta$ as well as $\alpha \beta$ from part b(i)

**CaptainBlack** · October 12th 2008, 11:05 PM

Originally Posted by maybeline9216

Please teach me how to do part b(iii).

lets assume we are going to do this without using the quadratic formula.

$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha \beta=(\alpha+\beta)^2-4\alpha \beta$

so:

$\alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha \beta}$

and we know that $\alpha+\beta=k=7$ and that $\alpha \beta=2k-3=11$

CB

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