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## #1 2018-03-13 11:38:34

Mathegocart
Member
Registered: 2012-04-29
Posts: 1,965

### Somewhat confused on a quadratic question.

Hi, amiable friends, and I took a quadratics test today. I was somewhat confused by this question.
If the coefficients of a quadratic equation are real, and if one of the roots is (1-sqrt(3)), the other root must be (1+sqrt(3)).
TRUE?
FALSE?
This shouldn't be true, should it? The other root can be anything, and the coefficents would still be real.

Last edited by Mathegocart (2018-03-13 11:39:06)

The integral of hope is reality.
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## #2 2018-03-13 17:59:51

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,220
Website

### Re: Somewhat confused on a quadratic question.

False: consider the equation
.

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## #3 2018-03-15 16:22:07

ganesh
Registered: 2005-06-28
Posts: 26,817

### Re: Somewhat confused on a quadratic question.

Hi,

False.

Other than

, there are any number of possibilities. Some are
.

is one of them.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #4 2018-03-15 21:58:40

Libera
Member
Registered: 2018-03-07
Posts: 16

### Re: Somewhat confused on a quadratic question.

ganesh wrote:

Hi,

False.

Other than

, there are any number of possibilities. Some are
.

is one of them.

I'd find interesting to share the different ways we use to approach the problem. My immediate view was graphical, for instance. And yours algebraic, I suppose.
I "saw" a parabola "free to move" but its fixed point

Last edited by Libera (2018-03-15 23:43:23)

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## #5 2018-03-15 23:06:13

Alg Num Theory
Member
Registered: 2017-11-24
Posts: 339
Website

### Re: Somewhat confused on a quadratic question.

Mathegocart wrote:

If the coefficients of a quadratic equation are real, and if one of the roots is (1-sqrt(3)), the other root must be (1+sqrt(3)).
TRUE?
FALSE?

If the coefficients were rational, then the statement would be true. It’s false if you allow the coefficients to be irrational, as the replies below your post show.

If the coefficients are real and one of the roots is complex (and non-real), then the other root must be the complex conjugate. This statement also fails if you allow the coefficients to be complex.

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## #6 2018-03-16 00:24:14

Alg Num Theory
Member
Registered: 2017-11-24
Posts: 339
Website

### Re: Somewhat confused on a quadratic question.

Alg Num Theory wrote:

If the coefficients were rational, then the statement would be true.

In general, if the coefficients are rational and one root is an irrational surd, then the other root must be the conjugate surd. This is not true if the coefficients can be irrational as well as rational.

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