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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,917

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.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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False: consider the equation .

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 24,640

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

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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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 189
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Mathegocart wrote:

If the coefficients of a quadratic equation are real, and if one of the roots is (1-sqrt(3)), the other root

mustbe (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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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 189
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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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