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Math Help - Prove no real roots

  1. #1
    Junior Member ibnashraf's Avatar
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    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.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ibnashraf View Post
    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
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    Junior Member ibnashraf's Avatar
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    i did use the discriminant and ended up with a^2+b^2 > 2ab .... from there i'm not sure what to do exactly
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ibnashraf View Post
    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}
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    Junior Member ibnashraf's Avatar
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    Thanks alot man ..... i got it out
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    Junior Member ibnashraf's Avatar
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    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?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ibnashraf View Post
    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
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  8. #8
    Moo
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    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 ^^
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    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
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  10. #10
    Moo
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    Quote Originally Posted by Jhevon View Post
    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
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