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Workout number of solutions for x^3 + 5 = 0

How would you the number of solutions for x^3+5 along with a reason for the answer please ?
Reply 1
Avatar for BabyMaths
BabyMaths
Original post by T-GiuR
How would you the number of solutions for x^3+5 along with a reason for the answer please ?


If x3+5=0x^3+5=0 then x3=−5x^3=-5. There is only one real number x such that x3=−5x^3=-5.
Original post by BabyMaths
If x3+5=0x^3+5=0 then x3=−5x^3=-5. There is only one real number x such that x3=−5x^3=-5.

Awesome thanks, so that would mean there is only 1 real solution or could that be imaginary as its -5 ?

Also for an x^2 - PI = 0, is there only 2 solutions ? X= +or- Root PI ?
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by T-GiuR
How would you the number of solutions for x^3+5 along with a reason for the answer please ?



I do not know the level of rigour you are looking but

f(x)=x3-5 is a cubic, so clearly at least one real solution as pointed out earlier
f'(x)=3x2>=0, so increasing and therefore no more real solutions
Original post by TeeEm
I do not know the level of rigour you are looking but

f(x)=x3-5 is a cubic, so clearly at least one real solution as pointed out earlier
f'(x)=3x2>=0, so increasing and therefore no more real solutions

All the question asks is "By sketching the curves y = f(x), state how many real solutions to the followingequations exist. Note, you do not have to find the solutions, but you must explain
clearly your reasoning."

So with that being said, there is only 1 solution ? but I dont know how to provide a reason ?
Reply 5
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davros
16
Original post by T-GiuR
All the question asks is "By sketching the curves y = f(x), state how many real solutions to the followingequations exist. Note, you do not have to find the solutions, but you must explain
clearly your reasoning."

So with that being said, there is only 1 solution ? but I dont know how to provide a reason ?


That suggests you just need to sketch the curve reasonably accurately and show that it only crosses the x axis at one point.
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by T-GiuR
All the question asks is "By sketching the curves y = f(x), state how many real solutions to the followingequations exist. Note, you do not have to find the solutions, but you must explain
clearly your reasoning."

So with that being said, there is only 1 solution ? but I dont know how to provide a reason ?


then you just need to sketch the graph of y=x^3-5 and etc etc ...
Original post by davros
That suggests you just need to sketch the curve reasonably accurately and show that it only crosses the x axis at one point.

Would that be the given reason ? Only intersects X-Axis once ?
Original post by T-GiuR
Would that be the given reason ? Only intersects X-Axis once ?
Not really. An intersection with the x-axis is a root, so if you said

"it only has one root because it only intersects the x-axis once"

it's the same as saying

"it only has one root because it only has one root".

which is not exactly convincing.

In this case I'd argue along the lines of x^3-5 is strictly increasing, so it can have at most one root.
Reply 9
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TeeEm
19
Original post by DFranklin
Not really. An intersection with the x-axis is a root, so if you said

"it only has one root because it only intersects the x-axis once"

it's the same as saying

"it only has one root because it only has one root".

which is not exactly convincing.

In this case I'd argue along the lines of x^3-5 is strictly increasing, so it can have at most one root.


I mentioned this in an earlier post but it appears this chap is at the very start of his C1 (objective: determining number of roots by graphical methods), no rigour needed:smile:
just find the graph of f(x) = x^3
Given that, the graph of f(x) = x^3 + 5 is the same graph, just translated vertically by +5
(edited 9 years ago)
Original post by DFranklin
Not really. An intersection with the x-axis is a root, so if you said

"it only has one root because it only intersects the x-axis once"

it's the same as saying

"it only has one root because it only has one root".

which is not exactly convincing.

In this case I'd argue along the lines of x^3-5 is strictly increasing, so it can have at most one root.

So if a question of the same kind, but says x^2 - PI = 0, would there be 2 solutions as a result of +or- ROOT PI ?
Original post by T-GiuR
So if a question of the same kind, but says x^2 - PI = 0, would there be 2 solutions as a result of +or- ROOT PI ?


There could be 2, 1 or no solutions (for quadratic equations), looking at the graphs of such equations should help quite a lot.
Original post by tim_123
There could be 2, 1 or no solutions (for quadratic equations), looking at the graphs of such equations should help quite a lot.

Thanks everyone :smile:
Reply 14
Avatar for davros
davros
16
Original post by T-GiuR
All the question asks is "By sketching the curves y = f(x), state how many real solutions to the followingequations exist. Note, you do not have to find the solutions, but you must explain
clearly your reasoning."




Original post by DFranklin
Not really. An intersection with the x-axis is a root, so if you said

"it only has one root because it only intersects the x-axis once"

it's the same as saying

"it only has one root because it only has one root".

which is not exactly convincing.

In this case I'd argue along the lines of x^3-5 is strictly increasing, so it can have at most one root.


Tbh I interpreted this question as "IF you're capable of sketching the graph of the function, THEN you can write down why there is only one (real) root", but without knowing what level the OP is working at it's difficult to say how much rigour/explanation is expected!

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