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# Help with A level maths? Watch

1. (Original post by Uni12345678)
I have no idea if this is correct or not, but...

EDIT : sorry just realised the last part is a little wrong , b squared - 4ac = -2 squared - 4(1)(-root10) so should be 4 + 4root10 so still 2 real roots but not 16 + 4root10
Wouldn't it be ? Then you continue while taking into account two cases in which it's either negative and positive roots of 10. The one you used shows 2 real roots and the other will show that the other 2 are imaginary.
2. Isn't it a necessity that the number of roots = the degree of the equation?
3. (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.
4. (Original post by RDKGames)
Wouldn't it be ? 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
5. (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?
6. (Original post by oShahpo)
The answer must obviously be 4 then, without needing to find the roots.
"How many real roots..."
7. (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)
8. (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.
9. (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
10. (Original post by Bath~Student)
Lovely - all correct.
Nope, might want to read more carefully next time.
11. (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?
12. (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.
13. (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.
14. (Original post by Zacken)
Nope, might want to read more carefully next time.
ahh, it's always the subtleties with me.
15. (Original post by Zacken)
Nope, might want to read more carefully next time.
Ahahahahha burn
16. (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 th degree polynomial to have roots in , it all falls out from being the algebraic closure of but the proof(s) uses relatively advanced concepts.
17. (Original post by Uni12345678)
Ahahahahha burn
That said, I like your method (once fixed) very much. Good on you for spotting it.
18. (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) ?
19. (Original post by Zacken)
That said, I like your method (once fixed) very much. Good on you for spotting it.
Thank you feel clever today loll💁🏻
20. (Original post by Uni12345678)
Oh right thanks so is that always the same as the total number of roots (including both real and imaginary) ?
Yes the degree equals the total number of roots; real + imaginary. But as Zacken stated, the coefficients need to be real, which they are in this case and many others.

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