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.
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 $\displaystyle x \in \mathbb{R}$
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 $\displaystyle a \ne b$
Hello Jhevon,
I'm a little confused ...
"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 $\displaystyle -4(a-b)^2$ ^^
the "if" was in front of the "a and b are unequal"
the way ibnashraf re-worded it, yes.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 ?
yes, i realize that. but i am a good little boy and i prove what i am told to proveActually, this implication is an equivalence, since the discriminant is $\displaystyle -4(a-b)^2$ ^^