Help with A level maths?

Isn't it a necessity that the number of roots = the degree of the equation?

(edited 7 years ago)

Reply 22

Original post by oShahpo

Isn't it a necessity that the number of roots = the number of solutions to the equation?

Since roots and solutions are more or less precisely the same thing, then yeah. Not sure why you're asking though.

Reply 23

Original post by RDKGames

Wouldn't it be

x^2(x-2)^2=10 \rightarrow x(x-2)=\pm\sqrt{10}

? Then you continue while taking into account two cases in which it's either negative root 10 and positive root 10? The one you used shows 2 real roots and the other will show that the other 2 are imaginary.

Yeah sorry I forgot to include negative root10, you're right
The answer is still 2 real roots because the other two would be imaginary
Thanks :smile:

Reply 24

Original post by Zacken

Since roots and solutions are more or less precisely the same thing, then yeah. Not sure why you're asking though.

The answer must obviously be 4 then, without needing to find the roots.
*EDIT: I meant the number of roots = the degree of the equation. Apologies.
Is the number of roots necessarily equal the degree of the equation?

Reply 25

Original post by oShahpo

The answer must obviously be 4 then, without needing to find the roots.

"How many real roots..."

Reply 26

Original post by oShahpo

Isn't it a necessity that the number of roots = the number of solutions to the equation?

Yes, in other words, how many times would the graph cross the x axis
There are different types of roots and generally only ask real roots rather than imaginary (like root negative 10 is imaginary because you cannot square root a negative number)

Original post by oShahpo

The answer must obviously be 4 then, without needing to find the roots.
*EDIT: I meant the number of roots = the degree of the equation. Apologies.
Is the number of roots necessarily equal the degree of the equation?

Yes, but the question asks for real roots so we need to take into account how many are imaginary and how many real out of all those that there are.

Reply 28

Original post by oShahpo

The answer must obviously be 4 then, without needing to find the roots.
*EDIT: I meant the number of roots = the degree of the equation. Apologies.
Is the number of roots necessarily equal the degree of the equation?

What is the degree of the equation? Sorry

Reply 29

Original post by Bath~Student

Lovely - all correct.

Nope, might want to read more carefully next time.

Reply 30

Original post by RDKGames

Yes, but the question asks for real roots so we need to take into account how many are imaginary and how many real out of all those that there are.

Oh, I see. Missed that bit. Thanks.
Btw, do you know the proof for this statement? Do you have a link or something?

Original post by Uni12345678

What is the degree of the equation? Sorry

Degree of the polynomial is the highest exponent. The question has a polynomial of degree order 4.

Reply 32

Original post by Uni12345678

Yes, in other words, how many times would the graph cross the x axis
There are different types of roots and generally only ask real roots rather than imaginary (like root negative 10 is imaginary because you cannot square root a negative number)

Sorry I mistyped, I fixed my comment though.

Reply 33

Bath~Student

2

Original post by Zacken

Nope, might want to read more carefully next time.

ahh, it's always the subtleties with me.

Reply 34

Original post by Zacken

Nope, might want to read more carefully next time.

Ahahahahha burn

Reply 35

Original post by oShahpo

Is the number of roots necessarily equal the degree of the equation?

Yes, as long as the coefficients of the polynomial are real, then the Fundamental theorem of Algebra forces every

n

th degree polynomial to have

n

roots in

\mathbb{C}

, it all falls out from

\mathbb{C}

being the algebraic closure of

\mathbb{R}

but the proof(s) uses relatively advanced concepts.

(edited 7 years ago)

Reply 36

Original post by Uni12345678

Ahahahahha burn

That said, I like your method (once fixed) very much. Good on you for spotting it.

Reply 37

Original post by RDKGames

Degree of the polynomial is the highest exponent. The question has a polynomial of degree order 4.

Oh right thanks so is that always the same as the total number of roots (including both real and imaginary) ?

Reply 38