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can a cubic have no real roots?

Thr answer to this seems simple, but is it possible to have a cubic which has no roots. The algebra sais it is, but when i think of it graphically, kt dosnt seem possible.

Reply 1

yusufu

16

its impossible for a cubic to have no real roots. what algebra are you referring to?

Reply 2

shrek

OP

1

well i was thinking that a cubic must have a discriminent just not quite sure what it is

Reply 3

yusufu

16

a cubic "starts" in one quadrant of the graph and "ends up" in the diagonally opposite quadrant. in order for this to happen it must cross y = 0, i.e. have a root.

Reply 4

meathead

5

You can have a cubic equation with no real roots, but it must have complex coefficients. For example, P(z) = (z - i)(z + i)(z - 2i) clearly has no real roots. However it isn't possible to plot such a cubic in the classic 2D 'x versus f(x)' way.

Obviously the only cubic you can plot in that way is one with real coefficients, and such a polynomial must have at least one real root, as complex roots of polynomials with real coefficients come in conjugate pairs (so it could have at most 2 complex roots). As you point out, this is obvious when you look at such a cubic equation graphically.

Reply 5

dvs

10

No, it's not possible. In fact any polynomial of odd degree MUST have at least 1 real root (since complex roots occur in conjugate pairs).

Edit:
Of course, as meathead point out, this only holds when we consider polynomials with real coefficients.

Reply 6

Kernel

1

Don't all polynomials with an odd highest power have at least 1 reat root?

Reply 7

barmuda

1

what is the real root of the equation 32x^3 240x^2 64x-384=0

Reply 8

Dysf(x)al

20

Original post by barmuda

what is the real root of the equation 32x^3 240x^2 64x-384=0

Dude, this thread is nearly 12 years old. Please make your own thread. next time.

Anyway, start by cancelling out common factors. All your coefficients are divisible by 16.

Reply 9

the bear

20

cubics do not have a discriminant.

you can write a cubic C as

C = L*Q where L in linear ...( x - a ) and Q is quadratic ( x² + bx + c )

Q certainly has a discriminant, which could be negative. But C will still have the root from L.... which is a.

Reply 10

anosmianAcrimony

20

Original post by meathead

You can have a cubic equation with no real roots, but it must have complex coefficients. For example, P(z) = (z - i)(z + i)(z - 2i) clearly has no real roots. However it isn't possible to plot such a cubic in the classic 2D 'x versus f(x)' way.

Obviously the only cubic you can plot in that way is one with real coefficients, and such a polynomial must have at least one real root, as complex roots of polynomials with real coefficients come in conjugate pairs (so it could have at most 2 complex roots). As you point out, this is obvious when you look at such a cubic equation graphically.

You are not a meathead - there's definitely a brain in there.

Reply 11

ghostwalker

17

This thread is 11 years old!

Reply 12

Nagini.Kill.

4

Original post by ghostwalker

This thread is 11 years old!

That's older than half of the TSR's user base

Reply 13

B_9710

18

Original post by the bear

cubics do not have a discriminant.

you can write a cubic C as

C = L*Q where L in linear ...( x - a ) and Q is quadratic ( x² + bx + c )

Q certainly has a discriminant, which could be negative. But C will still have the root from L.... which is a.

Cubics do have a discriminant in fact

I know this thread is 12 years old.

can a cubic have no real roots?

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