# Thread: Prove no real roots

1. ## Prove no real roots

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

2. Originally Posted by ibnashraf
Question: Prove that $(a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ 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

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

4. Originally Posted by ibnashraf
i did use the discriminant and ended up with $a^2+b^2 > 2ab$ .... from there i'm not sure what to do exactly
that's not what i got. i got that the discriminant is $\triangle = -4a^2 - 4b^2 + 8ab$

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

it is easier than you think. think "factorize". then recall that $[f(x)]^2 \ge 0$ for all $x \in \mathbb{R}$

5. Thanks alot man ..... i got it out

6. 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
$(a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ has no real roots."

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

7. 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
$(a^2+b^2)x^2 + 2(a+b)x + 2 = 0$ 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:

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

this is exactly what we proved. we assumed $a \ne b$ and then found the discriminant and showed that it has to be negative when $a \ne b$

8. 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 $(a^2+b^2)x^2 + 2(a+b)x + 2 = 0 \implies a \ne b$, doesn't it ?

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

9. Originally Posted by Moo
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"
the "if" was in front of the "a and b are unequal"

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

Actually, this implication is an equivalence, since the discriminant is $-4(a-b)^2$ ^^
yes, i realize that. but i am a good little boy and i prove what i am told to prove

10. 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
Ok ! Next time, I'll read the first post more carefully