Show that if r1 and r2 are distinct real roots

[merlin2007]

Show that if r1 and r2 are distinct real roots of x^2+px+8=0, then
abs(r1+r2)>4 * SQRT(2).

I'm trying to use the theory of quadratics:

r1+r2 = -p
r1*r2 = 8

From here, I thought I had to make the absolute value of p > 4 * SQRT(2), but I don't see how to proceed.

Thanks in advance.

 

[pka]

Use the discriminate! Because the roots are distinct and real we have:
\(\displaystyle \begin{array}{rcl}
p^2 - 4(1)(8) & > & 0 \\
p^2 & > & 32 \\
\left| p \right| & > & 4\sqrt 2 \\
\end{array}.\)

Using the sum of the roots: \(\displaystyle \begin{array}{rcl}
r_1 + r_2 & = & - p \\
\left| {r_1 + r_2 } \right| & = & \left| p \right| \\
\end{array}.\)

Do you see how to finish?

 

[merlin2007]

Thanks a lot

Thanks.

You know, I feel incredibly stupid that I didn't see that...but I guess that's why I'm practicing these problems.