# Locating Roots

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If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)

What about for f(x) = (2-x)

^{2}? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?
Last edited by Pyruvic Acid; 4 weeks ago

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If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)

**Pyruvic Acid**)If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)

^{2}? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?Anyway, a change in sign guarantees at least one root,

BUT the converse is not true; no change in sign does not guarantee that there are no roots. As per your example: (2-x)^2 has one root in interval [0,3] yet no change of sign. A change of sign is when you go from <0 to >0 or vice versa.

Last edited by RDKGames; 4 weeks ago

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(Original post by

For a continuous f(x) the intermediate value theorem says that if there is a change in sign of f between two points, there is AT LEAST one root between the two points.

Anyway, a change in sign guarantees at least one root,

BUT the converse is not true; no change in sign does not guarantee that there are no roots. As per your example: (2-x)^2 has one root in interval [0,3] yet no change of sign. A change of sign is when you go from <0 to >0 or vice versa.

**RDKGames**)For a continuous f(x) the intermediate value theorem says that if there is a change in sign of f between two points, there is AT LEAST one root between the two points.

Anyway, a change in sign guarantees at least one root,

BUT the converse is not true; no change in sign does not guarantee that there are no roots. As per your example: (2-x)^2 has one root in interval [0,3] yet no change of sign. A change of sign is when you go from <0 to >0 or vice versa.

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**Pyruvic Acid**)

If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)

^{2}? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?

At a given root of a polynomial, you will get a sign change if, and only if, the exponent of the corresponding factor is odd.

E.g in your example, there is a root at x=2. The corresponding factor is (x-2)^2. The exponent is 2 (i.e. even) so no sign change.

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4th attempt, having realised my previous ones were not quite correct and needed more qualifying than was useful.

At a given root of a polynomial, you will get a sign change if, and only if, the exponent of the corresponding factor is odd.

E.g in your example, there is a root at x=2. The corresponding factor is (x-2)^2. The exponent is 2 (i.e. even) so no sign change.

**ghostwalker**)4th attempt, having realised my previous ones were not quite correct and needed more qualifying than was useful.

At a given root of a polynomial, you will get a sign change if, and only if, the exponent of the corresponding factor is odd.

E.g in your example, there is a root at x=2. The corresponding factor is (x-2)^2. The exponent is 2 (i.e. even) so no sign change.

^{2}+7x+12 in the interval [-4,-2]? Although the exponent of corresponding factor is even, there is a change in sign, no?

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What about for x

**Pyruvic Acid**)What about for x

^{2}+7x+12 in the interval [-4,-2]? Although the exponent of corresponding factor is even, there is a change in sign, no?How do you know the corresponding factor is even?

Last edited by ghostwalker; 4 weeks ago

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(Original post by

What is it in factor format? (x-a)(x-b)

How do you know the corresponding factor is even?

**ghostwalker**)What is it in factor format? (x-a)(x-b)

How do you know the corresponding factor is even?

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