# P3 Qs.

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#1
Find the value of the constant k so that the polynomial P(x), where:
P(x)=x^2+kx+11
has a remainder 3 when it is divided by (x-2).
Show that, with thsi value of k, P(x) is positive for all real x.

I did the first part, but can't do the second. Here's what I did for the first bit:

x^2+kx+11=3
2^2+k(2)+11=3
2k=-12
k=-6

Also, (29/32)^1/5 = 1/2(29)^1/5

How?

How do you make 29/32 into just 29, with 1/2 outside the bracket?

Thanks a lot if you can help.
0
15 years ago
#2
(Original post by TheQueen1986)
Find the value of the constant k so that the polynomial P(x), where:
P(x)=x^2+kx+11
has a remainder 3 when it is divided by (x-2).
Show that, with thsi value of k, P(x) is positive for all real x.

I did the first part, but can't do the second. Here's what I did for the first bit:

x^2+kx+11=3
2^2+k(2)+11=3
2k=-12
k=-6

Also, (29/32)^1/5 = 1/2(29)^1/5

How?

How do you make 29/32 into just 29, with 1/2 outside the bracket?

Thanks a lot if you can help.
so do you mean you cant do the 'show the real roots' bit?
0
15 years ago
#3
(Original post by TheQueen1986)
Also, (29/32)^1/5 = 1/2(29)^1/5

How?

How do you make 29/32 into just 29, with 1/2 outside the bracket?

Thanks a lot if you can help.
(29/32)^1/5 = (29^1/5)/(32^1/5)

32^1/5 = 2 => (29/32)^1/5 = (1/2)29^1/5
0
15 years ago
#4
(Original post by Nylex)
(29/32)^1/5 = (29^1/5)/(32^1/5)

32^1/5 = 2 => (29/32)^1/5 = (1/2)29^1/5
u no the real roots bit yea? but its a negative answer so there are no real roots? 0
15 years ago
#5
(Original post by TheWolf)
lol ur right, I made a stupid mistake....
It was wrong I think, cos I didn't show P(x) was +ve for all real x.. I took off a 3 with the equation he had underneath.
0
15 years ago
#6
Ummm....if it is always positive, then it has no real roots. The discriminant is negative.
0
15 years ago
#7
(Original post by Nylex)
It was wrong I think, cos I didn't show P(x) was +ve for all real x.. I took off a 3 with the equation he had underneath.
ahh true then i was right in the end, stop confusin me 0
15 years ago
#8
Alternatively, complete the square
p(x) = (x-3)^2 + 2
squared terms are always greater or equal to 0, so p(x) is always positive.
0
15 years ago
#9
(Original post by JamesF)
Alternatively, complete the square
p(x) = (x-3)^2 + 2
squared terms are always greater or equal to 0, so p(x) is always positive.
how do u do it with the b^2 - 4ac method? i did it and its a negative number
0
15 years ago
#10
(Original post by TheWolf)
how do u do it with the b^2 - 4ac method? i did it and its a negative number
Exactly, it should. If it is positive, then it means that the graph crosses the x-axis - so goes negative at some point. If it is negative then there are no real roots, and so it doesnt cross the x-axis, therefore it is positive for all real x.
0
15 years ago
#11
(Original post by JamesF)
Exactly, it should. If it is positive, then it means that the graph crosses the x-axis - so goes negative at some point. If it is negative then there are no real roots, and so it doesnt cross the x-axis.
Show that, with thsi value of k, P(x) is positive for all real x.
does a negative answer proof that? i thought that only a positive answer show that it is positive for all real x
0
15 years ago
#12
(Original post by TheWolf)
ahh true then i was right in the end, stop confusin me You said real roots, that was what confused me.
0
15 years ago
#13
(Original post by TheWolf)
does a negative answer proof that? i thought that only a positive answer show that it is positive for all real x
Yes a negative answer proves that. A positive answer shows that it isnt always positive, because then there exists real roots.
0
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