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Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

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)

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zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

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Re: Somewhat confused on a quadratic question.

False: consider the equation

.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,385

Re: Somewhat confused on a quadratic question.

Hi,

False.

Other than

, there are any number of possibilities. Some are

.

is one of them.

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Libera

Member

Registered: 2018-03-07

Posts: 16

Re: Somewhat confused on a quadratic question.

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)

Alg Num Theory

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Registered: 2017-11-24

Posts: 693

Website

Re: Somewhat confused on a quadratic 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?

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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Me, or the ugly man, whatever (3,3,6)

Alg Num Theory

Member

Registered: 2017-11-24

Posts: 693

Website

Re: Somewhat confused on a quadratic question.

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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Me, or the ugly man, whatever (3,3,6)