IF $a>0$ and $b<0$, which of the following statements are true about the value of (x) that solve the eq0ution $x^2 - ax + b = 0$
a)they have opposite signs b)their sum is greater than zero c)their product equals $- b$
Now my choice was a) and c), but c) is incorrect and I'm not sure why since b is negative (<0) which means that the values of x will be opposite (a negative product) hence a), but why not c)? The answer is a) and b) but I'm not sure why b) is an answer. It COULD be I think, but I believe c) has to be. Thanks everyone.
$x^2-ax+b=0 \implies $
$x = \frac {a\pm\sqrt {a^2-4b}}2$
$b <0$ so $-4b>0$ so $\sqrt {a^2 -4b} >\sqrt {a^2}= a >0$.
So $a-\sqrt {a^2-4b} < a -a =0$.
So $\frac {a + \sqrt {a^2-4b} }2> 0 >\frac { a -\sqrt {a^2-4b}}2 $
so a)is true.
b) $\frac {a + \sqrt {a^2-4b}}2+ \frac {a - \sqrt {a^2-4b}}2 = a >0$ so b).
And $ \frac {a + \sqrt {a^2-4b}}2 \frac {a -\sqrt {a^2-4b}}2=\frac {a^2-(a^2-4b)}4=b \ne -b $ so not c)