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.

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- Aug 25th 2008, 04:57 PMibnashrafProve 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 PMJhevon
- Aug 25th 2008, 05:19 PMibnashraf
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 PMJhevon
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 PMibnashraf
Thanks alot man ..... i got it out (Nod)

- Aug 25th 2008, 05:56 PMibnashraf
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 PMJhevon
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 PMMoo
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 PMJhevon
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 ?

Quote:

Actually, this implication is an equivalence, since the discriminant is $\displaystyle -4(a-b)^2$ ^^

- Aug 25th 2008, 10:47 PMMoo