polynomial question

tangaloomaflyer

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Hi,

I'm not sure if this is a problem with my arithmetic or if I misunderstand the question. I'll write out the whole question with the answers to the first two parts, but it is the third part I am stuck on:

Given that p and q are roots of the equation 3x2 +x + 2 = 0

(i) evaluate 1/p2 + 1/q2


Answer: -11/4

working:
1/p2 +1/q2 = (p+q)2 - 2pq / (pq)2
Re-arrange 3x2 +x + 2 = 0 tothe form x2 - (p+q)x + (pq) = 0
p+q = 1/3

pq = 2/3
[(p+q)2 - 2pq] / (pq)2 = (1/9 - 4/3) / (4/9) = -11/4

(ii) Form an equation with roots 1/p2 and 1/q2

Answer: 4x2 +11x + 9 = 0Working:
Make new roots P and Q
Using the form x2 - (P+Q)x + (PQ) = 0 equivalent to x2 - (b/a)x + (c/a) = 0
P+Q = 1/p2 + 1/q2 =-11/4 =-b/a therefore b=11, a=4
(pq)2 = 4/9 from [(p+q)2 - 2pq] / (pq)2 = (1/9 - 4/3) / (4/9)PQ = 1/(pq)2 = 9/4 = c/a
therefore a=4, b=11, c=9
equation is 4x2 + 11x + 9 = 0


(iii) Show that 27p4 = 11p + 10

Answer: ???


I am assuming that p is the root of the original equation 3x2 +x + 2 = 0

From the discriminant of the quadratic formula, sqrt(b2 - 4ac) = sqrt(-23), therefore the roots are not real. When I find the values of 27p4 and 11p + 10 for the two possible roots I get different values. I think this is due to my poor fractional arithmetic of complex numbers because I keep getting different values! I am sure there must be a simpler way.

My workings (apologies as I don't know how to write the radical sign or fractions using code):

Find roots using quadratic formula:

p = [-1 + sqrt(23)i]/6 or [-1 - 1 sqrt(23)i]/6


Find 27p4

p2 = (1 + 2sqrt(23)i +23)/36 or (1 - 2sqrt(23)i +23)/36

p4 = (2sqrt(23)i +24)2 / 362 or (-2sqrt(23)i +24)2 / 362
= (96sqrt(23)i + 484) / 1296

27p4 = 2sqrt(23)i + 242/24

find 11p +10

11p = -11/6 + 11/6 sqrt(23)i or -11/6 - 11/6 sqrt(23)i

11p+ 10 does not equal 2sqrt(23)i + 242/24



Any ideas? The question is from a section of the textbook on quadratics, so I'm thinking I need to make the equation (p2)(27p2)= 11p +10 and use the answers of the first two parts to logically prove the equation, but I'm stuck. Unless the p is referring to a root of 4x2 + 11x + 9 = 0 ?
 
Hi,

I'm not sure if this is a problem with my arithmetic or if I misunderstand the question. I'll write out the whole question with the answers to the first two parts, but it is the third part I am stuck on:

Given that p and q are roots of the equation 3x2 +x + 2 = 0

(i) evaluate 1/p2 + 1/q2


Answer: -11/4

working:
1/p2 +1/q2 = (p+q)2 - 2pq / (pq)2
Re-arrange 3x2 +x + 2 = 0 tothe form x2 - (p+q)x + (pq) = 0
p+q = 1/3

pq = 2/3
[(p+q)2 - 2pq] / (pq)2 = (1/9 - 4/3) / (4/9) = -11/4

(ii) Form an equation with roots 1/p2 and 1/q2

Answer: 4x2 +11x + 9 = 0Working:
Make new roots P and Q
Using the form x2 - (P+Q)x + (PQ) = 0 equivalent to x2 - (b/a)x + (c/a) = 0
P+Q = 1/p2 + 1/q2 =-11/4 =-b/a therefore b=11, a=4
(pq)2 = 4/9 from [(p+q)2 - 2pq] / (pq)2 = (1/9 - 4/3) / (4/9)PQ = 1/(pq)2 = 9/4 = c/a
therefore a=4, b=11, c=9
equation is 4x2 + 11x + 9 = 0


(iii) Show that 27p4 = 11p + 10

Answer: ???


I am assuming that p is the root of the original equation 3x2 +x + 2 = 0

From the discriminant of the quadratic formula, sqrt(b2 - 4ac) = sqrt(-23), therefore the roots are not real. When I find the values of 27p4 and 11p + 10 for the two possible roots I get different values. I think this is due to my poor fractional arithmetic of complex numbers because I keep getting different values! I am sure there must be a simpler way.

My workings (apologies as I don't know how to write the radical sign or fractions using code):

Find roots using quadratic formula:

p = [-1 + sqrt(23)i]/6 or [-1 - 1 sqrt(23)i]/6


Find 27p4

p2 = (1 + 2sqrt(23)i +23)/36 or (1 - 2sqrt(23)i +23)/36

p4 = (2sqrt(23)i +24)2 / 362 or (-2sqrt(23)i +24)2 / 362
= (96sqrt(23)i + 484) / 1296

27p4 = 2sqrt(23)i + 242/24

find 11p +10

11p = -11/6 + 11/6 sqrt(23)i or -11/6 - 11/6 sqrt(23)i

11p+ 10 does not equal 2sqrt(23)i + 242/24



Any ideas? The question is from a section of the textbook on quadratics, so I'm thinking I need to make the equation (p2)(27p2)= 11p +10 and use the answers of the first two parts to logically prove the equation, but I'm stuck. Unless the p is referring to a root of 4x2 + 11x + 9 = 0 ?


When you square + sqrt(23)i you do not get 23! Yes, sqrt(23) squared is 23 but i^2 is not 1. What is i^2? What is the square of
+ sqrt(23)i?

BTW, this is not beginning algebra
 
The square of (sqrt23)i is -23

I posted in beginning algebra because I'm fairly certain it's just my arithmetic that is at fault.
 
When you square + sqrt(23)i you do not get 23! Yes, sqrt(23) squared is 23 but i^2 is not 1. What is i^2? What is the square of
+ sqrt(23)i?

BTW, this is not beginning algebra

I solved it in the end, it was my arithmetic that was the problem, which is why I posted in beginning algebra
 
Hint: Rewrite the first equation as
\(\displaystyle 3x^2\ =\ -(x+2)\)

It should be obvious what to do next.

I don't follow.

I would have to multiply the LHS by \(\displaystyle 9x^2\) but then the RHS would be a cubic equation.

Also, doesn't it re-arrange to \(\displaystyle 3x^2\ =\ -(x-2)\)?

My arithmetic is pretty terrible.


 
I keep posting replies to this post but they have not been appearing - I don't know why.

I have managed to solve this but i often end up with the wrong signs so i think it is just down to my arithmetic

Limax - I'm no sure where to go with this?

If I square both sides so that the term on the LHS is fourth power, then I end up with a quadratic on the RHS, but I can't get any further.
If I multiply both sides by 9x^2 then I end up with a cubic equation on the RHS.
 
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