Asked by Majek
Help. If p and q are the roots of the equation 2x^2-x-4=0. Find the equation whose roots are p-(q/p) and q-(p/q). Working please
Answers
Answered by
Reiny
I will use the sum and product of roots properties.
for the given :
sum of roots = p + q = 1/2
product of roots = pq = -4/2 = -2
new roots are p- q/p and q - p/q
sum of new roots = p - q/p + q - p/q
= p^2q + q^2p - q^2 - p^2
= pq(pq + qp) - ( (p+q)^2 - 2pq)
= -2(-4) - (1/4 + 4)
= 8 - 17/4
= 15/4
product of new roots
= (p- q/p)(q - p/q)
=pq - p^2/q - q^2/p + 1
= 1/2 + 1 - (p^3 + q^3)
aside:
(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3
(p+q)^3 = p^3 + q^3 + 3pq( p + q)
p^3 + q^3 = (p+q)^3 - 3pq( p + q)
= 1/8 - 3(-2)(1/2)
back to
1/2 + 1 - (p^3 + q^3)
= 3/2 - (1/8 - 3(-2)(1/2) )
= 3/2 - (-23/8) = 35/8
new equation:
x^2 - 15x/4 + 35/8 = 0
times 8
8x^2 - 30x + 35 = 0
I think you better check my algebra and arithmetic
for the given :
sum of roots = p + q = 1/2
product of roots = pq = -4/2 = -2
new roots are p- q/p and q - p/q
sum of new roots = p - q/p + q - p/q
= p^2q + q^2p - q^2 - p^2
= pq(pq + qp) - ( (p+q)^2 - 2pq)
= -2(-4) - (1/4 + 4)
= 8 - 17/4
= 15/4
product of new roots
= (p- q/p)(q - p/q)
=pq - p^2/q - q^2/p + 1
= 1/2 + 1 - (p^3 + q^3)
aside:
(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3
(p+q)^3 = p^3 + q^3 + 3pq( p + q)
p^3 + q^3 = (p+q)^3 - 3pq( p + q)
= 1/8 - 3(-2)(1/2)
back to
1/2 + 1 - (p^3 + q^3)
= 3/2 - (1/8 - 3(-2)(1/2) )
= 3/2 - (-23/8) = 35/8
new equation:
x^2 - 15x/4 + 35/8 = 0
times 8
8x^2 - 30x + 35 = 0
I think you better check my algebra and arithmetic
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