# Prove no real roots

• Aug 25th 2008, 04:57 PM
ibnashraf
Prove no real roots
Question: Prove that $\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots, if a and b are unequal.
• Aug 25th 2008, 04:59 PM
Jhevon
Quote:

Originally Posted by ibnashraf
Question: Prove that $\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots, if a and b are unequal.

I'd go with finding the discriminant. (you remember what that is, right?) show that it is negative if a and b are unequal and you are done
• Aug 25th 2008, 05:19 PM
ibnashraf
i did use the discriminant and ended up with $\displaystyle a^2+b^2 > 2ab$ .... from there i'm not sure what to do exactly
• Aug 25th 2008, 05:27 PM
Jhevon
Quote:

Originally Posted by ibnashraf
i did use the discriminant and ended up with $\displaystyle a^2+b^2 > 2ab$ .... from there i'm not sure what to do exactly

that's not what i got. i got that the discriminant is $\displaystyle \triangle = -4a^2 - 4b^2 + 8ab$

now we need to show that this is negative if $\displaystyle a \ne b$

it is easier than you think. think "factorize". then recall that $\displaystyle [f(x)]^2 \ge 0$ for all $\displaystyle x \in \mathbb{R}$
• Aug 25th 2008, 05:41 PM
ibnashraf
Thanks alot man ..... i got it out (Nod)
• Aug 25th 2008, 05:56 PM
ibnashraf
what confuses me though is the way the question was stated.

according to what we actually proved, the question should have been "Prove that a and b are unequal if
$\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots."

Rather what we actually proved is its converse i think .... what do u suggest?
• Aug 25th 2008, 05:59 PM
Jhevon
Quote:

Originally Posted by ibnashraf
what confuses me though is the way the question was stated.

according to what we actually proved, the question should have been "Prove that a and b are unequal if
$\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots."

Rather what we actually proved is its converse i think .... what do u suggest?

the question asked us to prove an implication:

$\displaystyle a \ne b \implies (a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots.

this is exactly what we proved. we assumed $\displaystyle a \ne b$ and then found the discriminant and showed that it has to be negative when $\displaystyle a \ne b$
• Aug 25th 2008, 10:43 PM
Moo
Hello Jhevon,

I'm a little confused (Blush)...

"a and b are unequal if ...=0 has no real roots" is the same as "if ...=0 has no real roots, (then) a and b are unequal"

And this last sentence rather leads to $\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0 \implies a \ne b$, doesn't it ?

Actually, this implication is an equivalence, since the discriminant is $\displaystyle -4(a-b)^2$ ^^
• Aug 25th 2008, 10:46 PM
Jhevon
Quote:

Originally Posted by Moo
Hello Jhevon,

I'm a little confused (Blush)...

"a and b are unequal if ...=0 has no real roots" is the same as "if ...=0 has no real roots, (then) a and b are unequal"

the "if" was in front of the "a and b are unequal"

Quote:

And this last sentence rather leads to $\displaystyle (a^2+b^2)x^2 + 2(a+b)x + 2 = 0 \implies a \ne b$, doesn't it ?
the way ibnashraf re-worded it, yes.

Quote:

Actually, this implication is an equivalence, since the discriminant is $\displaystyle -4(a-b)^2$ ^^
yes, i realize that. but i am a good little boy and i prove what i am told to prove (Sun)
• Aug 25th 2008, 10:47 PM
Moo
Quote:

Originally Posted by Jhevon
the "if" was in front of the "a and b are unequal"

the way ibnashraf re-worded it, yes.

yes, i realize that. but i am a good little boy and i prove what i am told to prove (Sun)

Ok ! Next time, I'll read the first post more carefully (Rofl)